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ADVANCED    COURSE 


IN 


ALGEBRA 


BY 


WEBSTER  WELLS,   S.B. 

PROFESSOR  OF  MATHEMATICS   IN  THE   MASSACHUSETTS 
INSTITUTE  OF  TECHNOLOGY 


BOSTON,  U.S.A. 

D.   C.   HEATH   &   CO.,   PUBLISHERS 

1904 


mVS 


Copyright,  1904, 
By  WEBSTER  WELLS. 


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PREFACE 

In  preparing  the  present  work,  the  author  has  endeavored 
to  meet  the  needs  of  Colleges  and  Scientific  Schools  of  the 
highest  rank. 

The  development  of  the  subject  follows  in  the  main  the 
author's  College  Algebra;  but  numerous  improvements  have 
been  introduced. 

Attention  is  especially  invited  to  the  following : 

1.  The  development  of  the  fundamental  laws  of  Algebra  for 
the  positive  and  negative  integer,  the  positive  and  negative 
fraction,  and  zero,  in  Chaps.  I  and  II. 

In  the  above  treatment,  the  author  has  followed  to  a  certain 
extent  The  Number  System  of  Algebra,  by  Professor  H.  B.  Fine ; 
who  has  very  courteously  permitted  this  use  of  his  treatise. 

2.  The  development  of  the  principles  of  equivalence  of 
equations,  and  systems  of  equations,  both  linear  and  of  higher 
degrees;  see  §§  116-123,  182,  233-6,  396,  442,  470,  477,  and 
478. 

3.  The  prominence  given  to  graphical  representation. 

In  Chap.  XIV,  the  student  learns  how  to  obtain  the  graphs 
of  linear  equations  with  two  unknown  numbers,  and  of  linear 
expressions  with  one  unknown  number.  He  also  learns  how 
to  represent  graphically  the  solution  of  a  system  of  two  linear 
equations,  involving  two  unknown  numbers,  and  sees  how  inde- 
terminate and  inconsistent  systems  are  represented  graphically. 

The  graphical  representation  of  quadratic  expressions,  with 
one  unknown  number,  is  taken  up  in  §  465 ;  and,  in  §  467,  the 
graphical  representation  of  equal  and  imaginary  roots. 

2S1763 


iv  PREFACE 

The  principles  are  further  developed  for  simultaneous  quad- 
ratics, in  §§  482  and  483;  and  for  expressions  of  any  degree, 
with  one  unknown  number,  in  §§  744  and  745. 

At  the  end  of  Chap.  XVIII,  the  student  is  taught  the  graphi- 
cal representation  of  the  fundamental  laws  of  Algebra  for  pure 
imaginary  and  complex  numbers. 

In  Chap.  XXXVII,  the  graphical  representation  is  given  of 
Derivatives  (§  751),  of  Multiple  Boots  (§  755),  of  Sturm's 
Theorem  (§  762),  and  of  a  Discontinuous  Function  (§  766). 

4.  In  Chap.  VII,  there  are  given  the  Remainder  and  Factor 
Theorems,  and  the  principles  of  Symmetry. 

5.  In  Chap  VIII  will  be  found  every  method  of  factoring 
which  can  be  done  advantageously  by  inspection,  including 
factoring  of  symmetrical  expressions.  In  this  chapter  is  also 
given  Solution  of  Equations  by  Factoring  (§  182). 

6.  In  the  earlier  portions  of  Chap.  XI,  the  pupil  is  shown 
that  additional  solutions  are  introduced  by  multiplying  a 
fractional  equation  by  an  expression  which  is  not  the  L.C.M. 
of  the  given  denominators ;  and  is  shown  how  such  additional 
solutions  are  discovered. 

7.  In  §§  264  and  265,  the  student  is  taught  how  to  find  the 

values  of  expressions  taking  the  indeterminate  forms  ^j  — > 

Ox  00,  and  oo  —  oo. 

8.  All  work  coming  under  the  head  of  the  Binomial  Theo- 
rem for  positive  integral  exponents  is  taken  up  in  the  chapter 
on  Involution. 

9.  In  developing  the  principles  of  Evolution,  all  roots  are 
restricted  to  their  principal  values. 

10.  In  the  examples  of  §  398,  the  pupil  is  taught  to  reject 
all  solutions  which  do  not  satisfy  the  given  equation,  when 
the  roots  have  their  principal  values. 

11.  The  development  of  the  theory  of  the  Irrational  Num- 
ber, and  its  graphical  representation  (§§  399-406). 

12.  The  development  of  the  fundamental  laws  of  Algebra 
for  Pure  Imaginary  and  Complex  Numbers  (Chap.  XVIII). 


PREFACE  V 

13.  The  use  of  the  general  form  ax^  +  6a;  +  c  =  0,  in  the 
theory  of  quadratic  equations  (§§  454-6). 

14.  The  discussion  of  the  maxima  and  minima  values  of 
quadratic  expressions  (§  461). 

15.  The  chapter  on  Convergency  and  Divergency  of  Series 
(Chap.  XXVI). 

16.  In  Chap.  XXVIII  is  given  Euler's  proof  of  the  Binomial 
Theorem,  for  any  Eational  Exponent. 

17.  The  solution  of  logarithmic  equations  (§  604). 

18.  The  proof  of  the  formula  for  the  number  of  permuta- 
tions of  n  different  things,  taken  r  at  a  time  (§  624). 

19.  The  chapter  on  Theory  of  Numbers  (Chap.  XXXV). 

20.  In  the  chapter  on  Determinants,  the  double-suffix  nota- 
tion is  used  only  in  demonstrations  which  would  not  otherwise 
be  complete. 

Each  demonstration  of  a  general  principle  is  preceded  by 
an  illustration  showing  the  truth  of  the  principle  for  a  deter- 
minant of  the  third  order. 

Multiplication  of  determinants  is  taken  up  only  for  deter- 
minants of  the  second  and  third  orders. 

21.  In  Chap.  XXXVII  will  be  found  Symmetrical  Eunctions 
of  the  Eoots  (§  721) ;  a  shorter  proof  of  Descartes'  Rule 
(§  735) ;  improved  methods  for  finding  limits  to  the  roots 
(§§  739,  740);  the  demonstration  of  two  theorems  used  in  the 
proof  of  Sturm's  Theorem  (§  757) ;  and  a  discussion  of  Con- 
tinuous Functions  (§  765). 

22.  Chap.  XXXVIII  contains  the  solution  of  Cubic  Equa- 
tions by  Trigonometry,  in  Cardan's  Irreducible  Case  (§  788) ; 
also,  an  improved  discussion  of  Newton's  Method  for  deter- 
mining incommensurable  roots  (§  801). 

The  examples  and  problems  have  been  selected  with  great 
care,  and  include  many  varieties  not  found  in  the  College 
Algebra;  no  example  is  a  duplicate  of  any  in  the  College 
Algebra. 

The  manuscript  was  read  in  the  most  careful  manner  by 
Professor  George  D.  Olds,  of  Amherst  College,  who   offered 


VI  PREFACE 

many  suggestions;  these  have  added  materially  to  the  value 
of  the  work. 

The  author  would  be  under  great  obligations  to  any  one  who 
will  bring  to  his  attention  any  error  which  may  be  found  in 
the  book. 

WEBSTER  WELLS. 

Boston,  1904. 


CONTENTS 


CHAPTER  PAGE 

I.  Definitions.     Notation.     Positive  Integers         .         .  1 

II.     Rational  Numbers 11 

III.  Addition  and  Subtraction  of  Algebraic  Expressions. 

Parentheses 23 

N     IV.  Multiplication  of  Algebraic  Expressions   ...  32 

V.  J^wisiON  OF  Algebraic  Expressions        ....  3^ 

VI.     Integral  Linear  Equations 46 

VII.  Special  Methods  in  Multiplication  and  Division       .  59 

"^    VIII.     Factoring 76 

IX.  Highest  Common  Factor.      Lowest  Common  Multiple  .  99 

-^          X.     Fractions   . 108 

XI.  Fractional  and  Literal  Equations       ....  122 

XII.     Simultaneous  Linear  Equations 133 

XIII.  Discussion  of  Linear  Equations 149 

XIV.  Graphical  Representation 166 

XV.     Involution  and  Evolution 173 

XVI.      lNEQUALmE8_ 206 

XVII.     Surds.     Theory  of  Exponents 212 

XVIII.  Pure  Imaginary,  and  Complex  Numbers        .         .         .  246 

XIX.     Quadratic  Equations 261 

XX.  Equations  solved  like  Quadratics        ....  292 

XXI.  Simultaneous  Quadratic  Equations       ....  298 

XXII.     Indeterminate  Linear  Equat10"ns 323 

v/.  XXIIL     Ratio  and  Proportion      . 328 

XXIV.     Variation 338 

XXV.     Progressions 343 

XXVI.  CONVERGENCY    AND   DIVERGENCY    OF   SeRIES        .            .            .  362 

\ 

vii 


VIU 


CONTENTS 


CIIAPTEK  PAGE 

XXVII.     Undetermined  Coefficients 378 

XXVIII.     The  Binomial  Theobem 392 

XXIX.     Logarithms  .         .        .         .' 399 

XXX.  Compound  Interest  and  Annuities  ....  422 

XXXI.  Permutations  and  Combinations        ....  428 

XXXII.     Probability           .         . 436 

"*  XXXIII.     Continued  Fractions ;  449 

XXXIV.     Summation  of  Series 459 

XXXV.     Theory  of  Numbers 471 

XXXVI.     Determinants 478 

XXXVII.     Theory  of  Equations 502 

XXXVIII.  Solution  of  Higher  Equations          .         .         .         ,  545 

Appendix.  Cauchy's  Proof  that  every  Equation  has  a  Root  576 

Answers . 


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ALGEBRA 


I.    DEFINITIONS.    NOTATION.    POSITIVE 
INTEGERS 

1.  In  Algebra,  the  operations  of  Arithmetic  are  abridged  and 
generalized  by  means  of  Symbols. 

SYMBOLS  REPRESENTING  NUMBERS 

2.  The  symbols  usually  employed  to  represent  numbers  are 
the  Arabic  Numerals,  and  the  Letters  of  the  Alphabet. 

The  numerals  denote  known  or  determinate  numbers. 

The  letters  denote  numbers  which  may  have  any  values 
whatever,  or  numbers  whose  values  are  to  be  determined. 

Numbers  occupying  similar  relations  in  the  same  investiga- 
tion are  often  represented  by  the  same  letter,  distinguished 
by  different  accents;  as  a',  a",  a'",  read  "  a  prime /^  "a  second/^ 
"  a  third,"  etc. 

They  may  also  be  distinguished  by  different  subscript  num- 
bers; as  tti,  ttg,  ttg,  read  "a  sub  one,"  "a  sub  two,"  "a  sub  thi-ee" 
etc. 

SYMBOLS  REPRESENTING  OPERATIONS 

3.  The  Sign  of  Addition,  -f ,  is  read  '^plus." 
The  Sign  of  Subtraction,  — ,  is  read  ^' minus." 

The  Sign  of  Multiplication,  x,  is  read  "times,"  "into,"  or 
"  multiplied  by." 

A  point  is  sometimes  used  instead  of  the  sign  x ;  thus, 
2.3.4  signifies  2x3x4. 

The  Sign  of  Division,  ^^  is  read  "  divided  by." 


2  ADVANCED  COURSE   IN  ALGEBRA 

SYMBOLS  OF  RELATION 

4.  The  Sign  of  Equality,  =,  is  read  "equals,^^  or  'Hs  equal  to." 
The  sign  =^  is  sometimes  used  for  the  words  "  is  not  equal  to.'''' 

The  Signs  of  Inequality,  >  and  <,  stand  for  "is  greater 
than'^  and  "is  less  than,'^  respectively. 

The  signs  ;5^  and  5C  are  sometimes  used  for  the  words  "  is  not  greater 
than''''  and  "is  not  less  than^''"'  respectively. 

SYMBOLS  OF  ABBREVIATION 

5.  The  Signs  of  Aggregation,  the  parentheses  (  ),  the  brackets 
[  ],  the  braces  j  \,  and  the  vmculiiyn  ,  indicate  that  what 
is  enclosed  by  them  is  to  be  taken  as  a  whole. 

The  Sign  of  Deduction,  .-.,  is  read  "therefore"  or  "hence." 
The  Sign  of  Continuation,  •••,  is  read  "and  so  on." 

THE  POSITIVE  INTEGER 

6.  By  the  number  of  things  in  a  group,  we  mean  that  attri- 
bute of  the  group  which  remains  unchanged  however  the  group 
may  be  changed,  provided  no  thing  is  divided  into  two  or 
more  things,  and  that  two  or  more  things  are  not  merged  into 
a  single  thing. 

That  is,  the  number  of  things  in  a  group  is  independent  of 
their  character,  of  the  order  in  which  they  may  be  arranged, 
and  of  the  way  in  which^they  may  be  associated  in  smaller 
groups. 

7.  The  numbers  of  things  in  two  groups  are  said  to  be  equal 
when  for  every  thing  in  either  group  there  is  a  thing  in  the 
other. 

8.  The  number  of  things  in  one  group  is  said  to  be  greater 
than  the  number  in  another,  or  the  number  in  the  second  group 
less  than  the  number  in  the  first,  when  for  every  thing  in  the 
second  group  there  is  a  thing  in  the  first,  but  not  one  in  the 
second  for  every  thing  in  the  first.    .    . 


POSITIVE  INTEGERS  3 

9.  The  Positive  Integer. 

We  define  a  positive  integer  as  the  number  of  things  in  a 
group. 

A  positive  integer  is  also  called  a  whole  number. 

To  ensure  generality  in  the  results,  we  represent  numbers  by- 
letters. 

In  the  remainder  of  the  present  chapter,  the  letters  a,  h,  c, 
etc.,  will  be  understood  as  representing  positive  integers. 

10.  If  a  and  h  stand  for  the  numbers  in  any  two  groups 
(that  is,  for  any  two  positive  integers),  we  use  the  statement 

a  =  6 
to  signify  that  the  numbers  are  equal  (§  7). 
The  statement  a  =  &  is  called  an  Equation. 

Again,  we  use  the  statements 

a>  b,  and  a<b 
to  signify  that  the  number  in  the  first  group  is  greater  or  less, 
respectively,  than  the  number  in  the  second  (§  8). 

These  statements  are  called  Inequalities. 

ADDITION  OF  POSITIVE  INTEGERS 

11.  Let  two  or  more  groups  contain  a,h,c,  •  •  •  things,  respec- 
tively. 

If  the  second  group  be  joined  to  the  first,  we  represent  the 
number  in  the  resulting  group  hj  a-{-h. 

If  to  the  latter  group  the  third  group  be  joined,  we  repre- 
sent the  number  in  the  resulting  group  by  a  +  h  -\-  c\  and 
so  on. 

After  all  the  groups  have  been  united  in  a  single  group,  the 
number  in  the  latter  group  is  expressed 

a  +  h-{-c-\ . 

This  result  is  called  the  Sum  of  the  positive  integers  a,  6,  c, 
etc. 

The  operation  of  finding  the  sum  is  called  Addition. 


4  ADVANCED  COURSE   IN   ALGEBRA 

12.  The  Commutative  and  Associative  Laws  for  Addition. 

« 
Addition  of  positive  integers  is  subject  to  the  following  laws: 

I.  The  Commutative  Law. 

To  add  6  to  a  is  the  same  as  to  add  a  to  h. 
Expressed  in  symbols, 

a-\-h  =  h  -\-  a. 

II.  Tlie  Associative  Law. 

To  add  the  sum  of  h  and  c  to  a  is  the  same  as  to  add  h  to  a, 
and  then  add  c  to  the  result.  / 

Expressed  in  symbols, 

a  -\-  (h  -\-  c)  =  a  -{-h  -\-  c. 

To  indicate  the  addition  of  6  +  c,  it  must  be  enclosed  in  parentheses 
(§5)- 

The  Commutative  and  Associative  Laws  follow  from  §  6; 
for  the  number  of  things  in  the  sum-group  is  independent  of 
the  order  in  which  they  may  be  arranged,  and  of  the  way  in 
which  they  may  be  associated  in  smaller  groups. 

The  Commutative  and  Associative  Laws  evidently  hold  for  the  sum  of 
any  number  of  positive  integers. 

MULTIPLICATION  OF  POSITIVE  INTEGERS 

13.  Finding  the  sum  of  b  positive  integers,  each  equal  to  a, 
is  called  multiplying  a  by  h. 

The  result  is  expressed  a  x  &,  or  simply  ah ;  thus, 

ah —  a -\-  a -\-  '•'  to  h  terms. 

The  sign  of  multiplication  is  usually  omitted  in  Algebra,  except  between 
Arabic  numerals. 

We  call  a  the  Multiplicand,  and  h  the  Multiplier. 

If  ah  be  multiplied  by  another  positive  integer,  c,  the  result 
is  expressed  ahc ;  and  so  on. 

If  any  number  of  positive  integers  be  multiplied  together, 
the  result  is  called  their  Product. 

The  operation  of  finding  the  product  is  called  Multiplication, 


POSITIVE  INTEGERS  5 

14.  The  Commutative,  Associative,  and  Distributive  Laws  for 
Multiplication. 

Multiplication  of  positive  integers  is  subject  to  the  following 
laws: 

I.  Tlie  Commutative  Law. 

To  multiply  a  by  6  is  the  same  as  to  multiply  h  by  a. 
Expressed  in  symbols,      ah  =  ha. 

II.  The  Associative  Law. 

To  multiply  a  by  the  product  of  h  by  c,  is  the  same  as  to 
multiply  a  by  h,  and  then  multiply  the  result  by  c. 
Expressed  in  symbols,  a(6c)  =  abc. 

III.  The  Distributive  Law. 

To  multiply  a  by  the  sum  of  h  and  c  is  the  same  as  to  multi- 
ply a  by  6,  and  then  a  by  c,  and  add  the  results. 
Expressed  in  symbols,  a(h  -\-  c)  =  ah  -{■  ac. 

15.  Proof  of  the  Commutative  Law. 

Let  there  be,  in  the  figure,  a  units  in  each  row,  and  h  rows. 
We  may  find  the  entire  number  of  units   by 
multiplying  the  number  in  each  row,  a,  by  the 

number  of  rows,  6. 

1111 
Thus,  the  entire  number  of  units  is  a6. 

1111 
We  may  also  find  the  entire  number  by  multi- 


plying the  number  in  each  vertical  column,  6,  by 

the  number  of  columns,  a.  rows. 

Thus,  the  entire  number  of  units  is  ba. 

Therefore,  ah = ha. 

16.   Proof  of  the  Associative  Law. 

By  the  definition  of  §  13, 

abc  =  ah  -{-  ah  -{-  •••  to  c  terms 

=  (a  +  a  -f  •••  to  h  terms)  +  (a  -j-  a  +  •••  to  &  terms) 

4-  •••  to  c  terms 
=  a  +  a  +  •  •  •  to  6c  terms, 

by  the  Associative  Law  for  Addition  (§  1 2) 
=  a(hc),  by  the  definition  of  §  13. 


6  ADVANCED  COURSE  IN  ALGEBRA 

17.  Proof  of  the  Distributive  Law. 
By  the  definition  of  §  13, 

a  (6  +  c)  =  a  +  a  4-  •  •  •  to  (b  +  c)  terms 

=  (a-\-a+  •••  to  b  terms)  -{-(a-^a-^  •••  to  c  terms), 
by  the  Associative  Law  for  Addition  (§  12), 
=  ab-{-  ac,  by  the  definition  of  §  13. 

18.  We  will  now  show  that  the  Commutative  and  Associa- 
tive Laws  for  Multiplication  hold  for  the  product  of  any 
number  of  positive  integers. 

We  will  first  prove  the  Commutative  Law  for  the  product  of 
three  positive  integers,  a,  b,  and  c. 

By  §  14,  II,     abc  =  a(bc)  =  {cb)a,  by  §  14,  I, 
=  cba  (§13). 

In  like  manner,  we  may  prove  abc  equal  to  the  product  of  a, 
b,  and  c  in  any  other  order. 

19.  We  will  now  prove  the  Associative  Law  for  the  product 
of  four  positive  integers,  a,  b,  c,  and  d. 

By  §  14,  I,  a(bcd)  =  (bcd)a  =  (be)  da  =  a{bc)d  (§  18) 

=  \_a(bc)Y  =  (abc)d  (§  14,  II)  =  abed. 

By  continuing  the  foregoing,  the  Commutative  and  Associa- 
tive Laws  may  be  proved  for  the  product  of  any  number  of 
positive  integers. 

The  Distributive  Law  for  Multiplication  holds  for  the  sum  of  any 
number  of  positive  integers,  as  is  evident  from  the  nature  of  the  demon- 
stration in  §  17. 

SUBTRACTION    OF    POSITIVE    INTEGERS 

20.  We  define  Subtraction  as  the  process  of  finding  one  of 
two  positive  integers  (the  Remainder),  when  their  sum  (the 
Minuend)  and  the  other  positive  integer  (the  Subtrahend)  are 
given. 

Thus,  subtraction  is  the  inverse  of  addition. 

21.  The  remainder,  when  b  is  subtracted  from  a,  is  expressed 
a  —  b. 


POSITIVE   INTEGERS  7 

Since,  by  the  definition  of  §  20,  the  sum  of  the  remainder 
and  the  subtrahend  equals  the  minuend,  we  have 

{a-h)^h  =  a.  (1) 

22.  If  a -\-  c  =  h  -\-  c,  then  a  =  h. 

For  if  the  numbers  of  units  in  the  sums  a-\-  c  and  6  +  c  are 
equal,  the  result  of  subtracting  the  units  in  c  from  each  sum 
will  be  the  same ;  that  is,  a  —  b. 

23.  It  follows,  precisely  as  in  §  22,  that 
If  a-\-c>h-[-Cj  then  a > 6. 
If  a-\-c<h-\-c,  then  a<6. 

24.  Rules  for  Subtraction. 

The  following  rules,  together  with  the  laws  of  §§12  and  14, 
are  sufficient,  if  suitably  combined,  to  determine  the'  result  of 
any  operation  with  positive  integers,  involving  only  addition, 
subtraction,  and  multiplication : 

(1)  a  —  {h-\-c)  =  a  —  'b  —  c. 

(2)  a  —  h  —  c     =a  —  c  —  b. 

(3)  a  -  (6  -  c)  =  a  -  6  4-  c. 

(4)  a  +  6  —  6    =  a. 

(5)  a-\-(b-c)  =  a-\-b-c. 

(6)  a-{-b  —  c     =a  —  c  +  b. 

(7)  a(b  —  c)       =ab  —  ac. 

25.  Proofs  of  the  Rules  for  Subtraction. 

Proof  of  (1). 

If  we  add  c,  and  then  6,  to  a  —  6  —  c,  or  (§  12,  I),  if  we  add 
b,  and  then  c,  or  (§  12,  II),  if  we  add  6  +  c,  the  result  is  a. 

That  is,  a  —  &  —  c  4-  (&  +  c)  =  a. 

Regarding  a  as  the  minuend,  6  +  c  as  the  subtrahend,  and 
a  —  6  —  c  as  the  remainder,  we  have 

a—  {b-\-c)  =  a  —  b  —  c. 


/ 


Hgj^ 


8  ADVANCED   COURSE   IN   ALGEBRA 

Proof  of  (2). 

By(l),     a-c-h  =  a-{c-irh)  =  a-(h-\-c)   (§12,1) 

=  a  -  6  -  c,  by  (1). 
Proof  of  (3). 
By§21,(l),  a-6  +  c  =  a-[(6-c)  +  c]+c 

=  a-(b-c)-c  +  c,  by  (1), 

=  a  —  (6  —  c). 
Proof  of  (4). 

We  have  a  +  b  —  b-{-b  =  a  +  b. 

Then,  by  §  22,  a  +  6  -  6  =  a. 

Proo/  of  (5). 

By  §21,        a  +  &-c  =  a  +  [(&-c)+c]-c 

=  a  +  (6-c)  +  c-c  (§12,11) 

=  a  +  (6-c),by(4). 
Proo/  0/  (6). 

By  §  12,  I,    a-\-b-c  =  b  +  a-c  =  b  + (a -c),  by  (5),    ^ 

=  (a  -  c)  +  &  (§  12, 1)  =  a  -  c  +  &. 
Proof  of  (7). 

By  §  21,  ab  —  ac  =  a  [(6  —  c)  +  c]  —  ac 

=  a(b-c)  +  ac-ac{^  14,  III)=a(6-c),  by  (4). 

It  is  important  to  observe  that  the  results  of  §  24  are  simply 
formal  consequences  of  §§  12, 14,  21,  and  22;  they  must  follow 
from  these  whatever  meaning  is  attached  to  the  symbols,  a,  b, 
c,  +>  —J  and  =. 

26.  Equations  (2)  and  (6),  §  24,  show  that  a  set  of  subtrac- 
tions, or  of  additions  and  subtractions,  can  be  performed  in  any 
order. 

Equation  (4)  shows  that  addition  is  the  inverse  of  subtraction. 

Equations  (1),  (3),  and  (5),  with  §  12,  II,  give  complete 
associative  laws  for  addition  and  subtraction. 

Equation  (7),  with  §  14,  III,  give  a  complete  distributive 
for  multiplication. 


POSITIVE  INTEGERS  9 

DIVISION  OF  POSITIVE  INTEGERS 

27.  We  define  Division  of  positive  integers  as  the  process  of 
finding  one  of  two  positive  integers  (the  Quotient),  which  when 
multiplied  by  another  positive  integer  (the  Divisor),  gives  a 
third  positive  integer  (the  Dividend). 

Thus,  division  is  the  inverse  of  multiplication. 

28.  The  quotient  when  a  is  divided  by  b  is  expressed  a-i-b, 

a 
or-. 

Since,  by  the  definition  of  §  27,  the  product  of  the  quotient 
by  the  divisor  gives  the  dividend,  we  have 

(1) 


to  c  terms,  we  must 


©'■ 

=  a. 

29.  If                       ac 

=  be,  then  a 

=  b. 

For,  if  a  +  aH to  c 

terms  = 

=  6  +  6+... 

have  a  =  &. 

30.   Rules  for.  Division. 

^  ^    b'  d~bd 

For  by  §  14,  I  and  II, 

a     c 
b'  d' 

bd  = 

<r' 

= 

:  ac,  by 

Also, 

ac 
bd' 

bd  = 

:  ac,  by 

Then,  by  §  29, 

a 
b 

c  _ 
'  d~ 

ac 
'bd 

ir') 


a 
/9x    ^      ad 
^^)    'c-Vc' 

d  a 

For  by  §28,(1),  l  '  l^l  ^^^ 

d 


10        ADVANCED  COURSE  IN   ALGEBRA 

AT       -u     /■i\  dd      c      ad     c      a  fd     c\   ^      e  ^  ^    tt 

Also,  by  (1),  —  •  -  =  7  •  -  •  -  =  -f  -  •  -  ,  by  §  14,  II, 

-^  ^  ^    be     d      b     G     d      b\G     dj     \  ' 


-W^'^^^- 

But  by  §  28, 

^  .  cd  =  dc  =  lx  cd. 
cd 

Whence,  by  §  29,      . 

^^=l,and^.^=^. 
cd                   be     d      b 

From  (A)  and  (B),  by  §  29,   -=|^. 

d 


(4) 


b     d  bd 

a      c      ad  ~  be 


(B) 


/o\  a     c__ad+J)c 

^  ^  b     d~~bd 

For  by  §  14,  I,  II,  and  III, 

=  acZ  +  6c,  by  §  28,  (1).  (C) 

Also,  ^:^L±h  .  M  =  ad  +  be,  by  §  28,  (1).  (D) 

oa 

From  (C)  and  (D),  by  §  29,   ^  +  ^  =  ^^  +  ^^- 


b      d         bd 
This  is  proved  in  the  same  manner  as  (3). 
The  results  of  §  30  are  simply  formal  consequences  of  §§  14, 
24,  28,  and  29 ;  and  must  follow  from  these  whatever  meaning 

is  attached  to  the  symbols  a,  b,  c,  +,—,=,  ab,  and  -• 


31.   Ifad  =  6c,tlien     f  -  •  6  jd  =  6  -  •  d  ,  by  §  28,  (1). 

Then,  by  §  14, 1  and  II,  -fbd\  =  ^  .  6  .  d  =  -fbd\ 

Then,  by  §29,  ^  =  ^-. 

b     d 


RATIONAL   NUMBERS  11 


II.    RATIONAL   NUMBERS 

32.  The  proofs  in  §  25  hold  only  when  the  result  of  every 
indicated  subtraction  is  ^.  positive  integer ;  for  the  laws  of  §§  12 
and  14  have  only  been  proved  for  the  case  in  which  all  the 
letters  involved  represent  positive  integers. 

A  result  like  (1)  has,  at  present,  no  meaning  unless  h  -\- c 
is  <  a ;  and  in  the  remainder  of  the  work  we  should  be  com- 
pelled to  limit  every  subtraction  to  cases  where  it  was  arith- 
metically possible. 

Unless,  then,  subtraction  is  to  be  very  much  restricted,  we 
must  consider  the  cases  where  the  subtrahend  equals,  or  is 
greater  than,  the  minuend ;  this  leads  to  the  introduction  into 
Algebra  of  Zero  and  the  Negative  Number. 

SYMBOLIC  EQUATIONS 

33.  The  equation  (1),  §  21,  is  not  an  equation  as  defined  in 
§  10,  unless  a  and  h  are  positive  integers,  and  a>b. 

But  if  we  agree  to  define  an  equation  as  simply  a  statement 
that  tico  symbols,  or  combinations  of  symbols,  are  of  such  a  char- 
acter that  one  m,ay  be  substituted  for  the  other  in  any  operation, 
then  (1),  §  21,  may  be  an  equation  whatever  the  values  of 
a  and  b. 

In  this  symbolic  definition  of  an  equation,  it  is  unnecessary 
that  there  should  be  any  real  things  to  which  the  symbols 
correspond. 

We  shall  attach  this  meaning  to  every  equation  throughout 
the  remainder  of  the  work  which  does  not  express  the  equality 
of  two  positive  integers. 

34.  Symbolic  Subtraction. 

If  we  regard  equation  (1),  §  21,  as  defining  a  —  b,  whatever 
the  values  of  a  and  b,  we  have  in  this  way  a  symbolic  definition 
of  subtraction  which  holds  universally. 


12  ADVANCED   COURSE  IN  ALGEBRA 

This  defines  subtraction  in  terms  of  symbolic  addition;  for 
the  sign  -\-  cannot  indicate  numerical  addition,  unless  the 
symbols  which  it  connects  are  positive  integers. 

It  is  perfectly  logical  to  define  an  operation  by  means  of 
an  equation. 

ZERO  AND  THE  NEGATIVE  INTEGER 

35.  In  determining  the  definitions  and  rules  of  operation  of 
zero  and  the  negative  integer,  we  make  the  assumption  that  the 
results  of  §§  12,  14,  22,  and  23  hold  for  these  symbols. 

If  all  the  letters  do  not  represent  positive  integers,  the 
results  of  §§  12,  14,  22,  and  23  are  regarded  as  symbolic 
statements. 

36.  Since  the  results  of  §  24  are  simply  formal  consequences 
of  §§  12, 14,  and  22,  and  the  definition  of  subtraction,  it  follows 
from  §  35  that  they  hold  for  the  above  symbols. 

If  the  results  of  §  24  do  not  have  a  positive  integral  inter- 
pretation, they  are  regarded  as  symbolic  statements. 

In  this  way  they  become  definitions  of  symbolic  addition, 
subtraction,  and  multiplication,  and  their  relations. 

37.  Zero. 

Every  letter  in  §§  37  to  42,  inclusive,  will  be  understood  as  representing 
a  positive  integer. 

Putting  6  =  a,  in  (1),  §  28,  we  have 

(a  —  a)  -j-  a  =  a.  (1) 

The  symbol  a  —  a,  if  a  is  any  positive  integer,  is  represented 
by  the  symbol  0,  called  zero. 

Then  (1)  becomes  0  +  a  =  a.  (2) 

Since  the  Commutative  Law  for  Addition  (§  12)  is  assumed 
to  hold  if   either  letter  equals  0  (§  35),  we  may  write  equa- 

^^o^  (2),  a  +  0  =  a.  (3) 

Again,  by  definition,  a  —  0  means  a  symbol  such  that  when 
0  is  added  to  it,  the  sum  is  a. 
That  is,  (a  —  0)  -f-  0  =  a. 


RATIONAL   NUMBERS  13 

Then,  by  (3),  a  -  0  =  a.  (4) 

Again,  by  the  definition  of  0, 

axO  =  a(b-b)  =  ab-  ab,  by  §  24,  (7), 

=  0,  by  definition.  (5) 

We  can  use  §  24,  (7),  in  the  above  proof ;  for  we  know  from  §  36  that 
it  holds,  even  if  the  result  does  not  have  a  positive  integral  interpretation. 

From  (5),  by  the  Commutative  Law  for  Multiplication, 

0  X  a  =  0.  (6) 

38.  The  Negative  Integer. 

Let  b  be  greater  than  a ;  and  suppose  b  =  a-{-  d,  where  d  is 
a  positive  integer. 

Then,  by  the  definition  of  subtraction,  b  —  a  =  d. 

Then,     a  —  b  =  a  —  (a  +  d)  =  a  —  a  —  d,hj^24:,  (1), 

=  0  —  d,  by  the  definition  of  0. 

We  can  use  §  24,  (1),  in  the  above,  for  we  know  that  it  holds,  even  if 
the  result  does  not  have  a  positive  integral  interpretation. 

We  then  define  a  —  6,  if  &  is  >  a,  as  being  equal  to  0  —  d. 
It  is  usual  to  write  —  d  instead  of  0  —  d 

Thus,  0-d  =  -d.  (1) 

The  symbol  —  d  is  called  a  Negative  Integer ;  in  contradis- 
tinction, the  positive  integer  d  may  be  written  +  d. 

39.  The  signs  +  and  — ,  when  used  in  the  above  manner, 
are  no  longer  signs  of  operation;  they  are  called  signs  of 
Affection^   Quality,  or  Opposition. 

If  no  sign  is  written,  the  sign  -}-  is  understood. 

40.  Rules  for  Addition,  Subtraction,  and  Multiplication,  involv- 
ing Negative  Integers. 

(1)    a-^(-b)  =  a-b. 

For,  by  the  definition  of  §  38, 

a  +  (-&)  =  «  +  (0  -  Z>)  =  a  +  0  -  &,  by  §  24,  (5), 
=  a-b,hy  ^  37,  (3). 


14  ADVANCED   COURSE   IN   ALGEBRA 

(2)  -a+(-b)=-(a  +  b). 

For,  —a+{—b)=  —  a+{0  —  b)=—a  +  0  —  b 

=  0  -  a  -  6,  by  §  12,  I, 
=  0  -  (a  +  ?^),  by  §  24,  (1), 
=  -  (a  +  ^>),  by  §  38. 

(3)  a-(-b)  =  a-\-b. 

For,       a  -  (-5)  =  a  -  (0  -  6)  =  tt  -  0  +  &,  by  §  24,  (3), 
=  a  +  6,  by  §37,(4), 

(4)  -a-(-b)  =  -a  +  b  =  b-a. 

For,  -  a  -  (-  6)  =  _  a  -  (0  -  6)  =  -  a  -  0  +  &=  -  a+6-0 

=  -«  +  &,  by  §37,  (4), 

=  6  -  a,  by  §  12,  I. 
Putting  b  for  a  in  (1)  and  (4),  we  have 

(5)  6-f  (-&)  =  0. 

(6)  -j)^b  =  0. 

(7)  _^,_(_6)=0. 

(8)  a(-6)  =  -a6. 

By  §  37,  (5),  0  =  a  X  0  =  a[6  +  (-  &)],  by  (5), 

=  o6 +  «(-&),  by  §  14,111. 
Then,  ab  —  ab  =  ab  -\-a(—  b). 

Whence,  by  §  22,  -  a6  =  a  (-  6). 

(9)  (-b)a  =  -ab  =  -ba. 

This  follows  from  (8)  by  §  14,  I. 

(10)  (-a)  xO  =  0. 

For,  by  the  definition  of  0, 

(- a)  X  0  =  (-  a) (b-b)  =  (-a)b-  (- a)b,  by  §  24,  (7), 

=  (_a6)-(-a6),by(9), 

=  0,by(7). 

(11)  Ox(-a)  =  0. 

This  follows  from  (10)  by  §  14,  I. 


RATIONAL  NUMBERS  15 

(12)  (-a)(-b)  =  ab. 

By   (10),  0=(-a)x0  =  (-a)[&  +  (-&)],by(5), 

=  (-  a)  6  +  (-  a)  (-  b),  by  §  14,  III, 
=  _a&4-(-a)(-6),  by  (9). 

Then  by  (6),   -  a&  +  a&  =  - a6  +  (- a)  (- 6). 

Then  by  §  22,  ^6  =  (-  a)  (-  6). 

41.  If  6  is  >  a,      0  +  6  >  0  +  a,  by  §  37,  (2). 
Then,  _a  +  a  +  6>-5  +  &  +  a,  by§40,  (6). 

Then  by  §23,  -a>-b.  (1) 

In  like  manner,  if  6  is  <  a,  —  a  is  <  —  b. 
These  may  be  regarded  as  defining  greater  and  less  inequality 
*in  negative  integers. 

42.  If  a  and  b  are  positive  integers, 
a<a-{-b;  or,  0  +  a  <  a  +  6,  by  §  37,  (2). 

Then,  by  §  23,  0  <  b. 

Then,  _  6  +  &  <  &  +  0,  by  §§  37,  (3),  and  40,  (6). 

Whence,  by  §  23,     -b< 0. 

SYMBOLIC  DIVISION 

43.  It  is  important  to  observe  that  the  result  of  §  29  does 
not  hold  when  c  =  0 ;  for  by  §  37,  (5),  a  x  0  =  6  x  0,  when  a  and 
b  are  not  equal. 

It  follows  from  this  that  the  proofs  in  §§  30  and  31  do  not 
hold  iib  =  0  or  d  =  0. 

44.  Symbolic  Division.     The  Fraction. 

The  proofs  in  §§  30  and  31  hold  only  when  the  result  of 
every  indicated  division  is  a  positive  integer. 

A  result  like  that  of  §  30,  (1),  has,  at  present,  no  meaning 

unless  -,  -,  and  —  are  positive  integers. 
b    d  bd 

But  if  we  regard  equation  (1),  §  28,  as  defining  -,  what- 

6 
ever  the  values  of  a  and  b,  provided  b  is  not  0  (§  43),  we  have 
a  symbolic  definition  of  division,  which  holds  universally. 


16  ADVANCED   COURSE   IN  ALGEBRA 

The  symbol  -,  with,  the  above  meaning  of  a  and  &,  is  called 
a  Fraction. 

The  symbols  -  and  -  are  considered  in  Chap.  XIII. 

If  a  and  h  are  positive  integers,  and  -  is  not  a  positive 
integer,  -  is  called  a  Positive  Fraction,  and  —-a  Negative 
Fraction. 

45.  We  make  the  assumption  that  the'  results  of  §§  12,  14, 

and  22  hold  for  the  symbol  -  as  defined  in  §  44;  whence,  it 

h 

follows  that  the  results  of  §  24  hold  for  the  symbol  -• 

b 
We  also  assume  that  the  result  of  §  29  holds  for  all  the  sym- 
bols considered  in  the  present  chapter,  provided  h  is  not  0. 

46.  Since  the  results  of  §§  30  and  31  are  simply  formal  con- 
sequences of  §§  14,  24,  and  29,  and  the  definition  of  division, 
it  follows  from  §  45  that  they  hold,  provided  h  and  d  are  not  0, 
even  if  the  results  do  not  have  a  positive  integral  interpretation. 

In  this  way,  the  results  of  §  30  become  definitions  of  addition, 

subtraction,  multiplication,  and  division,  for  the  symbol  -• 

47.  Since  the  results  of  §§  12,  14,  22,  and  24  hold  for  any 
of  the  symbols  considered  in  the  present  chapter  (§§  35,  36,  45), 
the  results  of  §§  37  and  40  hold  for  any  of  these  symbols;  for 
they  are  simply  formal  consequences  of  §§  12,  14,  22,  and  24, 
and  the  definition  of  subtraction. 

48.  Since  the  symbolic  definition  of  division  (§  44)  holds 
for  any  values  of  the  letters  involved,  we  have 

Again,  (12),  §  40,  holds  when  we  replace  a  by  -  (§  47). 

h 

Then,  /_2V_6)  =  ^.6  =  a.  (B) 


a            a 
-b          b 

Again, 

(7)— 

And  by  (9),  §  40, 

(-1)-(S")- 

—  a 

From  (C)  and  (D), 

—  a         a 
b            b 

Also,                      f- 

^)(- .)=-.. 

And  by  (8),  §  40, 

|(-.)=-g..)=. 

-a. 

From  (E)  and  (F), 

—  a     a 
-b      b 

|^/  RATIONAL  NUMBERS  17 

From  (A)  and  (B),  by  §  29,  which  is  supposed  to  hold  for 
the  symbol  -  6  (§  45), 

(1) 
(C) 

(P) 

(2) 
(E) 
(T) 

(3) 

The  results  (1),  (2),  and  (3)  hold  for  any  values  of  the  letters, 
provided  b  is  not  0. 

49.  Consider  the  equation  a6  =  0 ; 

where  a  and  b  may  be  positive  integers,  or  any  of  the  symbols 
considered  in  the  present  chapter. 
By  §  37,  (2),  whatever  the  value  of  c, 

0  +  ac  =  ac. 

Putting  ab  for  0,  we  have 

ab  -{-  ac  =  ac,  ot  a(b  -\-  c)  =  ac,  by  §  14,  III. 

Then  by  §  29,  if  a  is  not  0  (compare  §  43), 

&  +  c  =  c,  or  &  =  0  (§  22). 

Therefore,  either      a  =  0,  or  else  6  =  0. 

50.  It  is  advantageous,  at  this  point,  to  consider  the  nature 
of  the  argument  which  has  been  developed. 


18        ADVANCED  COURSE  IN  ALGEBRA 

In  Chap.  I,  we  defined  the  positive  integer,  and  the  opera- 
tions of  Addition,  Subtraction,  Multiplication,  and  Division 
with  positive  integers ;  and  we  showed  how  the  fundamental 
laws  of  §§  12,  14,  22,  and  29  followed  from  the  definitions  of 
Addition  and  Multiplication,  and  the  results  of  §§  24  and  30 
from  the  above  general  laws,  and  the  definitions  of  Subtraction 
and  Division. 

In  Chap.  II,  we  assumed  the  fundamental  laws  of  §§  12,  14, 
22,  and  29,  and  the  symbolic  definitions  of  Subtraction  and 
Division,  to  hold  universally ;  and  from  these  assumptions,  we 
derived  the  definitions  of  zero,  the  negative  integer,  and  the 
positive  and  negative  fraction,  and  the  rules  for  their  opera- 
tion, the  assum]3tions  being  just  sufficient  to  determine  these 
meanings  without  ambiguity. 

51.   Rational  Numbers. 

In  the  present  chapter,  we  have  considered  four  symbols  — 
zero,  the  negative  integer,  and  the  positive  and  negative  frac- 
tion—  which  are  subject  to  the  same  rules  as  positive  integers. 

The  result  of  every  operation  involving  only  addition,  sub- 
traction, multiplication,  and  division  —  whether  performed  on 
positive  integers,  or  on  the  symbols  themselves  —  can  be  ex- 
pressed either  as  a  i)ositive  integer,  or  as  one  of  the  symbols. 

For  this  reason,  we  shall  regard  these  symbols  as  numbers,; 
and  we  shall  term  the  entire  system  of  positive  and  negative 
integers,  and  positive  and  negative  fractions.  Rational  Numbers. 

Zero,  the  negative  integer,  and  the  positive  and  negative 
fraction,  are  essentially  artificial  numbers,  in  contrast  to  the 
natural  numbers  (positive  integers)  considered  in  Chap.  I. 

It  must  be  clearly  understood  that  they  are  simply  symbols 
for  the  results  of  operations  on  actual  groups  of  things,  which 
cannot  be  expressed  in  positive  integers.     (Compare  §  6.) 

We  shall  use  the  term  positive  number^  in  Chaps.  II  to  XVII,  inclusive, 
to  denote  a  positive  integer  or  a  positive  fraction  ;  and  the  term  negative 
number  to  denote  a  negative  integer  or  a  negative  fraction.  The  term 
number^  without  a  qualifying  adjective,  will  be  understood  as  signifying 
&  positive  or  negative  integer,  or  &  positive  or  negative  fraction. 


RATIONAL   NUMBERS  19 

Every  letter  will  be  understood  as  representing  a  positive  or  negative 
integer,  or  a  positive  or  negative  fraction,  unless  the  contrary  is  stated. 

52.  It  is  important  to  observe  that  the  results  of  §§24  and 
30,  and  all  the  results  of  Chap.  II,  follow  from  the  funda- 
mental laws  of  §§  12,  14,  22,  23,  and  29,  and  the  symbolic 
definitions  of  subtraction  and  division,  entirely  irrespective  of 
ivhether  or  no  the  symbols  have  any  numerical  meaning. 

Thus,  all  the  results  of  the  text  hold  for  any  symbols  which 
satisfy  the  fundamental  laws,  no  matter  what  their  meaning. 

53.  The  absolute  value  of  a  number  is  the  number  taken 
independently  of  the  sign  affecting  it. 

Thus,  the  absolute  value  of  —  3  is  3.. 

54.  The  results  of  §  42  hold  when  b  is  any  positive  number. 
Hence,  zero  is  less  than  any  positive  number,  and  any  negative 

number  is  less  than  zero. 

55.  Again,  by  §  47,  the  result  (1),  §  41,  also  holds  when  a 
and  b  are  positive  fractions. 

Hence,  of  tico  negative  numbers,  that  is  the  greater  which  has 
the  smaller  absolute  value. 

56.  Any  two  magnitudes  which  are  opposite  to  each  other  may 
be  represented  by  positive  and  negative  numbers,  in  Algebra. 

Thus,  in  financial  transactions,  we  may  represent  assets  by 
the  sign  ■-}-,  and  liabilities  by  the  sign  — ;  thus,  the  statement 
that  a  man's  assets  are  —  $  100,  means  that  he  has  liabilities 
to  the  amount  of  $  100. 

Again,  we  may  represent  motion  along  a  straight  line  in  a 
certain  direction  by  the  sign  +,  and  in  the  opposite  direction 


by  the 

!  sign 

—  ;  and  so  on. 

57. 

Graphical  Representation  of  Positive  and  Negative  Numbers. 

-8     - 

1 

-7     -6 

1         1 

-5      -4     -3     -2     -1         0      +1     +2     +8     +4     +5 

1          1          1          1          1          1          1          1          I          1          1 

4-6     +7    +8 

1          1          1 

//'      G'     F'      E'      ir      C     B'     A'       O     A        B       G       D       K      F       O       H 

The  entire  series  of  positive  and  negative  numbers  may  be 
represented  by  the  above  scale,  in  which  the  divisions  are  one 
unit  in  length. 


20  ADVANCED   COURSE   IN   ALGEBRA 

Distances  measured  to  the  right  of  0  represent  positive 
numbers,  and  to  the  left  of  0,  negative. 

Every  positive  or  negative  fraction  will  be  represented  by 
the  distance  from  0  to  a  point  between  two  consecutive  scale- 
marks. 

Thus,  the  number  —  3|  will  be  represented  by  the  distance 
from  0  to  a  point  two-thirds  the  way  from  C  to  U. 

58.  If  any  number,  positive  or  negative,  be  denoted  by  the 
symbol  a,  —a  will  represent  a  number  of  the  same  absolute 
value,  but  opposite  sign. 

It  follows  from  this  that  -f-  (—  a)  and  —  (—  a)  signify  num- 
bers of  the  same  absolute  value  as  —  a,  and  of  the  same  sign, 
and  opposite  sign,  respectively. 

That  is,       -f  (—  a)  =  —  a,  and  —  (—  a)  =  +  a.  (1) 

Similarly,   —  (+  a,)  =  —  a,  and  -f  (+  a)  =  +  a.      •  (2) 

From  (1)  and  (2), 

-f  (—  a)  =  -  (+  a),  and  -  (-  a)  =  -f-  (+  «)• 

DEFINITIONS 

59.  Addition,  Subtraction,  Multiplication,  and  Division  of 
any  algebraic  numbers  are  expressed  in  the  same  manner  as  in 
§§  11,  13,  20,  and  28. 

For  example,  2  ab  signifies  2  x  a  x  6. 

60.  If  a  number  be  multiplied  by  itself  any  number  of 
times,  the  product  is  called  a  Power  of  the  number. 

An  Exponent  is  a  number  written  at  the  right  of,  and  above 
another  number,  to  indicate  what  power  of  the  latter  is  to  be 
taken;  thus, 

a^,  read  " a  square,''  or  " a  second  power,"  denotes  a  x  a; 
a^,  read  "  a  cube,'"  or  "  a  third  power,"  denotes  a  x  a  x  a ; 
a*,  read  "  a  fourth,"  or  "  a  fourth  power,"  denotes  a  x  a  x  a  x  a ; 
and  so  on. 
If  no  exponent  is  expressed,  the  first  power  is  understood. 


RATIONAL   NUMBERS  21 

Thus,  a  is  the  same  as  a}. 

61.  Algebraic  Expressions. 

An  Algebraic  Expression,  or  simply  an  Expression,  is  a  number 
expressed  in  algebraic  symbols ;  as, 

2,  a,  or  2ic2-3a6+5. 

A  Monomial,  or  Term,  is  an  expression  whose  parts  are  not 

separated  by  the  signs  +  or  —  ;  as  2  a?^,   —  3  ah,  5,  or  — 

n 
A  monomial  is  sometimes  called  a  simple  expression. 

2  07^,  —  3  ah,  and  +  5  are  called  the  terms  of  the  expression 
2  a;2  -  3  a6  +  5. 

A  Positive  Term  is  one  preceded  by  a  plus  sign ;  as  +  5  a. 
Fot  this  reason  the  sign  +  is  often  called  the  positive  sign. 
If  no  sign  is  expressed,  the  term  is  understood  to  be  positive ; 
thus,  2  x^  is  the  same  as  -{-2  3?. 

A  Negative  Term  is  one  preceded  by  a  minus  sign ;  as  —  3  ah. 
For  this  reason  the 'sign  —  is  often  called  the  negative  sign  ; 
it  can  never  be  omitted  before  a  negative  term. 

A  Polynomial  is  an  expression  consisting  of  more  than  one 
term ;  as  a  +  6,  or  2  aj^  —  3  icy  —  5  2/^. 

A  polynomial  is -also  called  a  multinomial^  or  a  compound  expression. 

A  Binomial  is  a  polynomial  of  two  terms ;  as  a  +  &• 

A  Trinomial  is  a  polynomial  of  three  terms )  as  a  +  6  —  c. 

62.  The  Numerical  Value  of  an  expression  is  the  result 
obtained  by  substituting  particular  numerical  values  for  the 
letters  involved  in  it,  and  performing  the  operations  indicated. 

Thus,   if    a  =  4,   6  =  3,   c  =  5,    and    d  =  2,    the  numerical  value  of 

^a-^  —  -d^  =  4.  X  4  +  ^-^-23  =  16  +  10-8  =  18. 
h  3 

63.  A  monomial  is  said  to  be  rational  and  integral  when  it 
is  either  a  number  expressed  in  Arabic  numerals,  or  a  single 
letter  with  unity  for  its  exponent,  or  the  product  of  two  or 
more  such  numbers  or  letters. 


22        ADVANCED  COURSE  IN  ALGEBRA 

It  is  also  said  to  be  rational  and  integral  when  it  can  be 
reduced  to  either  of  the  above  forms. 

Thus,  3  a^6^,  being  equivalent  to  3  -  a  -  a  -  b  -  b  -  b,  is  rational 
and  integral. 

A  polynomial  is  said  to  be  rational  and  integral  when  each 

3 

term  is  rational  and  integral ;  as  2  a^ ab  +  c^. 

4 

64.  If  a  term  has  a  literal  portion  which  consists  of  a  single 
letter  with  unity  for  its  exponent,  the  term  is  said  to  be  of  the 
first  degree. 

The  degree  of  any  rational  and  integral  monomial  (§  63)  is 
the  number  of  terms  of  the  first  degree  which  are  multiplied 
together  to  form  its  literal  portion. 

Thus,  2a  is  of  the  Jirst  degree;  5ab,  of  the  second  degree; 
Sa^b^,  being  equivalent  to  3  aabbb,  is  of  the  ffth  degree ;  etc. 

The  degree  of  a  rational  and  integral  monomial-  equals  the 
sum  of  the  exponents  of  the  letters  involved  in  it. 

Thus,  a6V  is  of  the  eighth  degree. 

The  degree  of  a  rational  and  integral  polynomial  is  the 
degree  of  its  term  of  highest  degree. 

Thus,  2a^b  — 3  G-\-d^  is  of  the  third  degree. 

65.  Homogeneous  Terms  are  terms  of  the  same  degree. 
Thus,  a^,  3  bh,  and  —  5  a^y  are  homogeneous. 

A  polynomial  is  said  to  be  homogeneous  when  its  terms  are 
homogeneous ;  as  a^  +  3  b^c  —  4  xyz. 

66.  An  Axiom  is  a  self-evident  truth. 
The  following  are  assumed  as  axioms : 
X.   Any  number  equals  itself. 

2.  Any  number  equals  the  sum  of  all  its  parts. 

3.  Any  number  is  greater  than  any  of  its  parts. 

4.  Two  numbers  which  are  equal  to  the  same  number,  or  to 
equal  numbers,  are  equal. 

5.  If  for  any  number  in  an  expression  an  equal  number  be 
substituted^  the  vahie  of  the  expression  is  not  changed. 


ADDITION  23 


III.    ADDITION  AND  SUBTRACTION  OF 
ALGEBRAIC  EXPRESSIONS.    PARENTHESES 

67.  Addition  of  Positive  and  Negative  Numbers. 
If  a  and  b  represent  any  positive  numbers, 

a-{-(-b)  =  a-b',  (1) 

for  by  §  47,  the  result  (1),  §  40,  also  holds  when  either  a  or  6 
is  a  positive  fraction. 

If  6  is  >  a,  a  —  b  =  —(b  —  a)', 

for  the  result  of  §  38  also  holds  when  either  a  or  6  is  a  positive 
fraction  (§  47). 

Then,  a-}- (-b)  =  - (b-a).  (2) 

From  (1)  and  (2),  we  have  the  following  rule : 
To  add  a  positive  and  a  negative  number,  subtract  the  smaller 
absolute  value  (§  53)  fro7n  the  greater,  and  place  before  the  result 
the  sign  of  the  number  having  the  greater  absolute  value. 

Thus,  5i  +  (  -  3i)  =  If;  2  +  (  -  5)  =  -  3. 

68.  Addition  of  Negative  Numbers. 

If  a  and  b  represent  positive  numbers,  we  have  by  §  40,  (2), 

(-a)  +  (-6)  =  -(a  +  &). 
We  then  have  the  following  rule : 

To  add  two  negative  numbers,  add  their  absolute  values,  and 
put  a  negative  sign  before  the  result. 

Thus,  (-lJ)  +  (-2^)=-3i|. 

69.  Addition  of  Monomials. 

The  sum  of  a  and  b  is  a-{-b',  and  by  §  40,  (1),  the  sum  of  a 
and  —bisa  —  b;  hence. 

The  addition  of  monomials  is  effected  by  uniting  them  with 
their  respective  signs. 


24  ADVANCED   COURSE  IN  ALGEBRA 

Thus,  the  sum  of  a,  —b,c,—  d,  and  —  e  is 
a  —  b-\-c  —  d  —  e. 

Since  the  Commutative  Law  for  Addition  holds  for  any- 
rational  numbers  (§§  47,  51),  the  terms  may  be  united  in  any 
order,  provided  each  has  its  proper  sign. 

70.  Definitions.  If  two  or  more  numbers  are  multiplied 
together,  each  of  them,  or  the  product  of  any  number  of  them, 
is  called  a  Factor  of  the  product. 

Thus,  a,  b,  c,  ab,  ac,  and  be  are  factors  of  the  product  abc. 

71.  If  a  number  be  expressed  as  the  product  of  two  factors, 
each  is  called  the  Coefficient  of  the  other. 

Thus  in  2  ab,  2  is  the  coefficient  of  a6;  2  a  of  6;  a  of  2  &;  etc. 

72.  If  one  factor  of  a  product  is  expressed  in  Arabic 
numerals,  and  the  other  in  letters,  the  former  is  called  the 
numerical  coefficient  of  the  latter. 

Thus  in  2  ab,  2  is  the  numerical  coefficient  of  ab. 

If  no  numerical  coefficient  is  expressed,  the  coefficient  unity 

is  understood. 

Thus,  a  is  the  same  as  1  a. 

By  §  47,  the  result  (9),  §  40,  also  holds  when  6  is  a  positive 

fraction. 

2 
That  is,  —  3  a  is  the  product  of  —  3  and  a,  and  —-ab  is  the 

2  ^ 

product  of  —  -  and  ab. 

Then,   —  3   is   the   numerical  coefficient  of  a  in  —  3  a,  and 

2  .  .  .2 

—  -is  the  numerical  coefficient  of  ab  in  —-ab. 

o  o 

Thus,  in  a  negative  term  (§  61),  the  numerical  coefficient 
includes  the  sign. 

73.  Similar  or  Like  Terms  are  those  which  either  do  not 
differ  at  all,  or  differ  only  in  their  numerical  coefficients ;  as 
2  x^y  and  —  7  x^y. 

Dissimilar  or  Unlike  Terms  are  those  which  are  not  similar ; 
as  3  x^y  and  3  xy"^. 


ADDITION  25 

74.  Addition  of  Similar  Terms. 

1.  Find  the  sum  of  5  a  and  3  a. 

Since  the  Distributive  Law  for  Multiplication  (§  14,  III) 
holds  for  any  rational  numbers  (§  47),  we  have 

5  a  4-  3  a  ==  (5  +  3)  a  =  8  a. 

2.  Find  the  sum  of  5  a  and  —  3  a. 

By  §72,     5a  +  (-3a)  =  5a  +  (-3)a   ^ 

=  [5  +  (_3)]a  '  (§14,111) 

=  2  a.  (§  67) 

3.  Find  the  sum  of  —  5  a  and  3  a. 

( -  5)  a  +  3  a  =  [( -  5)  +  3]  a  =  -  2  a.  (§67) 

4.  Find  the  sum  of  —  5  a  and  —  3  a. 

(_5)a  +  (-3)a=[(-5)4-(-3)]a  =  -8a.      (§68) 

Therefore,  to  add  tvjo  similar  terms,  find  the  sum  of  their 
numerical  coefficients  (§§  67,  68,  72),  and  affix  to  the  result  the 
common  letters. 

5.  Find  the  sum  of  2  a,  —  a,  'da,  —  12  a,  and  6  a. 

Since  the  additions  may  be  performed  in  any  order,  we  may 
add  the  positive  terms  first,  and  then  the  negative  terms,  and 
finally  combine  these  two  results. 

The  sum  of  2  a,  3  a,  and  6  a  is  11  a. 

The  sum  of  —  a  and  — 12  a  is  — 13  a. 

Hence,  the  required  sum  is  11  a  +  (— 13  a),  or  —2  a. 

6.  Add  3(a-6),  -2(a-h),  6(a-b),  and  -4(a-6). 
The  sum  of  S(a-b)  and  6(a-b)  is  9(a-b). 

The  sum  of  —  2  (a  —  6)  and  —  4  (a  —  6)  is  —  6  (a  —  b). 
Then,  the  result  is  [  9  +  (-  6)]  (a  -  b),  or  3  (a  -  b). 

75.  If  the  terms  are  not  all  similar,  we  may  combine  the 
similar  terms,  and  unite  the  others  with  their  respective 
signs  (§  69). 

Ex.  Required  the  sum  of  12  a,  —5  a;,  —Sy^,  —5  a,  Sx, 
and  —  3  a;. 


26       ADVANCED  COURSE  IN  ALGEBRA 

The  sum  of  12  a  and  —  5  a  is  7  a. 

The  sum  of  —^x,  S-a:,  and  —3  a;  is  0;  for  the  result,  (5), 
§  40,  holds  for  any  value  of  b  (§  47). 
Hence,  the  required  sum  is  7  a  —  3  y^. 

76.  A  polynomial  is  said  to  be  arranged  according  to  the 
descending  powers  of  any  letter,  when  the  term  containing  the 
highest  power  of  that  letter  is  placed  first,  that  having  the  next 
lower  immediately  after,  and  so  on. 

Thus,  x*  +  3oc^y-2  xY  +  3  xy^  -4.y^ 

is  arranged  according  to  the  descending  powers  of  x. 

The  term  —4  ?/*,  which  does  not  involve  x  at  all,  is  regarded  as  contain- 
ing the  lowest  power  of  x  in  the  above  expression. 

A  polynomial  is  said  to  be  arranged  according  to  the  ascend- 
mg  powers  of  any  letter,  when  the  term  containing  the  lowest 
power  of  that  letter  is  placed  first,  that  having  the  next  higher 
immediately  after,  and  so  on. 

Thus,  a;*  +  3a;3?/-2a^/-f-3a;/-4?/* 

is  arranged  according  to  the  ascending  powers  of  y. 

11.   Addition  of  Polynomials. 

It  follows  from  §  12,  II,  and  §  24,  (5),  that  the  addition  of 
polynomials  is  effected  by  uniting  their  terms  with  their 
respective  signs. 

1.   Required  the  sum  of 

6  a  —  7  a^,  3  a;^  —  2  a  +  3  2/^  and  2  a^  —  a  —  mn. 

We  set  the  expressions  down  one  underneath  the  other, 
similar  terms  being  in  the  same  vertical  column. 

We  then  find  the  sum  of  the  terms  in  each  column,  and 
write  the  results  with  their  respective  signs ;  thus, 

6a-7a^ 
-2a  +  3a;2_^3^3 

—    a-\-2Qi?  —  mn 

3a  —  2  a;^  4-32/^  —  mn. 


ADDITION.    SUBTRACTION  27 

2.   Add  4a;-3aj2-ll  +  5aj3,    12aj2  _  7  _  8a^- 15a;,     and 
U-\-(ja^-\-10x-9x\ 

It  is  convenient  to  arrange  each  expression  in  descending 
powers  of  x  (§  76)  ;  thns, 

5x^-    Sx^-^-    4a;-ll 
6a^_    9a;2  +  10«  +  14 


and 


3oi? 

-            X-       4:. 

3. 

Add 

9(a  +  6)- 

-8(6  +  c),      _3(6  +  c)- 

-7(c  +  a), 

4(c-f 

-a)- 

5(a  +  6). 

0 

9(a  +  6)  - 

■    8(6 +  c) 

- 

■    3(6  +  c)-7(c  + 

a) 

—  6(c 

t  +  6) 

+  4(c  +  a) 

4(a  +  6)- 

-ll{b-{-c)-S(c  + 

a). 

4. 

Add 

3     ,  2, 

4"  +  5''- 

loandia-|6  +  |o. 

!- 

i"-\' 

^»- 

¥*!' 

11       14^,   8  „ 


SUBTRACTION 


78.   Subtraction  of  Monomials. 

By  §  47,  the  result  (3),  §  40,  holds  for  any  values  of  a  and  6. 
Hence,  to  subtract  a  monomial,  we  change  its  sign  and  add  the 
result  to  the  minuend. 

1.    Subtract  5  a  from  2  a. 

Changing  the  sign  of  the  subtrahend,  and  adding  the  result 
to  the  minuend,  we  have 

2a-5a  =  2a  +  (-5a)=-Sa  (§74). 


28  ADVANCED  COURSE  IN  ALGEBRA 

2.  Subtract  —  2  a  from  5  a. 

5a  —  (—2a)  =  5a-\-2a  =  7a. 

3.  Subtract  5  a  from  —  2  a. 

—  2a  —  5a  =  —  7  a. 

4.  Subtract  —  5  a  from  —2  a. 

-2a-(-5a)  =  -2a  +  5a  =  3a. 

5.  From  —  23  a  take  the  sum  of  19  a  and  —  5  a. 

It  is  convenient  to  change  the  sign  of  each  expression  which  is 
to  be  subtracted^  and  then  add  the  results. 

We  then  have     —  23  a  —  19  a  +  5  a,  or  —  37  a. 

* 
79.   Subtraction  of  Polynomials. 

By  §  36,  (1)  and  (3),  §  24,  hold  for  any  values  of  the  letters. 

Hence,  to  subtract  a  polynomial,  we  chayige  the  sig^i  of.  each  of 

its  terms,  and  add  the  result  to  the  minuend. 

1.  Subtract  7ab^-9  a'b  +  8  6^  from  5a^-2a^b  +  4.  ab\ 

It  is  convenient  to  place  the  subtrahend  under  the  minuend, 
so  that  similar  terms  shall  be  in  the  same  vertical  column. 

We  then  change  the  sign  of  each  term  of  the  subtrahend, 
and  add  the  result  to  the  minuend ;  thus, 
5a3-2a26  +  4a&2 

-9a^6  +  7a?>^  +  8?>^ 
5a^-{-la^b-'6ab''-%b\ 

The  student  should  perform  mentally  the  operation  of  changing  the 
sign  of  each  term  of  the  subtrahend. 

2.  Subtract  the  sum  of  ^x^  —  %x-\-x^  and  5  —  a?-\-x  from 
6a^-7x-4. 

We  change  the  sign  of  each  expression  which  is  to  be  sub- 
tracted, and  add  the  results. 

Qx^  -7x-4: 

~x^-9x^  +  Sx 

+    x^  —    X  —  5 


SUBTRACTION.    PARENTHESES  29 

80.  By  §  78,  subtracting  +  a  is  the  same  thing  as  adding 
—  a,  and  subtracting  —  a  the  same  thing  as  adding  +  a. 

That  is,  —  (h-  a)  =  4-  (—  a),  and  —  (—  a)  =  +  (+  a). 

In  these  results,  the  signs  within  the  parentheses  are  signs  of  affection 
(§  39),  and  those  without  signs  of  operation. 

Comparing  the  results  with  those  of  §  58,  where  all  the  signs  are  signs  of 
affection,  we  see  that  the  signs  +  and  — ,  when  used  as  signs  of  affection, 
are  subject  to  the  same  laws  as  when  used  as  signs  of  operation. 

Thus  the  meaning  attached  to  the  signs  +  and  — ,  in  §  39,  is  consistent 
with  their  meaning  as  symbols  of  operation. 

PARENTHESES 

81.  Removal  of  Parentheses. 

It  follows  from  §  12,  II,  and  §  24,  (5),  that:  ' 

Parentheses  preceded  by  a  -\-  sign  may  he  removed  without 
changing  the  signs  of  the  terms  enclosed. 

Again,  it  follows  from  §  24,  (1)  and  (3),  that : 

Parentheses  preceded  by  a  —  sign  may  be  removed  if  the  sign 
of  each  term  enclosed  be  changed,  from  -{-  to  —,  or  from  —  to  -{-. 

The  above  rules  apply  equally  to  the  removal  of  the  brackets,  braces, 
or  vinculum  (§  5). 

It  should  be  noticed,  in  the  case  of  the  latter,  that  the  sign  apparently 
prefixed  to  the  first  term  underneath  is  in  reality  prefixed  to  the  vinculum. 

Thus,  +  a  —  &  and  —  a  —  b  are  equivalent  to  +  (a  —  6)  and  —  («  —  6), 
respectively. 

Parentheses  often  enclose  others ;  in  this  case  they  may  be 
removed  in  succession  by  the  rules  of  §  81. 

Beginners  should  remove  one  at  a  time,  commencing  with  the  inner- 
most pair  ;  but  after  a  little  practice  they  should  be  able  to  remove  several 
signs  of  aggregation  at  one  operation,  in  which  case  they  should  commence 
with  the  outermost  pair. 


Ex.   Simplify  4:X  —  \3x-]-{—2x  —  x-a)}. 
We  remove  the  vinculum  first,  then  the  parentheses,  and 
finally  the  braces. 


30  ADVANCED   COURSE  IN  ALGEBRA 


Thus,  4:X  —  l3x-{-(—2x  —  x  —  a)l 

=  4: X  —  \S X  -\-  (—  2 X  —  X  -{-  a)l 
=  4:X  —  [3 X  —  2 X  —  X  -{-  a] 
=  4:X  —  3x-\-2x-{-x  —  a==4:X  —  a. 

82.   Insertion  of  Parentheses. 

To  enclose  terms  in  parentheses,  we  take  the  converse  of  the 
rules  of  §  81. 

Any  number  of  terms  may  be  enclosed  in  parentheses  preceded 
by  a  -\-  sign,  without  changing  their  signs. 

Any  number  of  terms  may  be  enclosed  in  parentheses  preceded 
by  a  —  sign,  if  the  sign  of  each  term  be  changed,  from  +  to  — , 
or  from  —  to  -\-. 

Ex.  Enclose  the  last  three  terms  of  a  —  b  +  c  —  d  +  e  in 
parentheses  preceded  by  a  —  sign. 

Result,  a  —  b  —  (—c-{-d  —  e).- 

EXERCISE  I 

1.  Add5(«  +  6),  -4(a;-y),  -6(a  +  &),  S{x-y),  -7(x-y),  and 
8(a  +  b). 

2.  Add  7  w2  -  2p  -  8  w3,  5  n^  -  m^  and  3  xy  -  4  m'^  +  2  nK 

3.  Add  a  -  9  -  8  aM-  16  a^,  5  +  15  a3  -  12  a  -  2  a%  and  6  a2  -  10  a^ 
-f-  11  a  -  13. 

4.  AddU{x  +  y)  -  17 (y  +  z),  i(y  +  z)- Id^z  +  x),  and  -1{z  +  x) 
-Six  +  y). 

5.  Add^x-^y-^z,  -lx  +  y  +  ^z,^ud~'lx-^y  +  lz. 

6.  Subtract  the  sum  of  8(w  +  n)  and  -  15(m  +  n)  from  -  19  (w  +  n). 

7.  Subtract  Sb-6d- 10  c  +  7a  from  4  (Z  +  12  a  -  13  c  -  9  6. 

8.  Subtract  41  a:^  -  2  x^  +  13  from  15  a;^  +  x  -  18. 

9.  Subtract  -p--m--  n  from  -m--p--n. 

3^       2  4  5  7^      3 

10.  Subtract  the  sum  of  4  x'^  -  9  ^/^  +  6  ^2  and  2  a;^  +  8  y2  _  n  ^2  from 

7  X2  -  3  ?/2  -  5  5!2. 

11.  From  the  sum  of  2a  +  36-4c  and  3&  +  4c-5(?  subtract  the 
sum  of  5c-6d  —  7a  and  — 7d!  +  8a  +  9  6. 


PARENTHESES  31 


Simplify  the  following : 

12.   9m-  (3  n  +  {4  m  -  [w  -  6  ?7i]}  -  [w  +  7  w]). 


13.  2a  +  (-6  6-{3c  +  (-4&-6c  +  a)}]. 

14.  7  X  -  {-  6  X  -  {-  6  X  ~  [-  4  X  -  S  X  -  2]}). 


15.    6n-[Sn-(Sn  +  6)-{-6n-\-7n-  5}]. 


16.   4a-  [a-{-7a-(8a-5a  +  3)-(-6a-2a-9)}]. 


17.  x-{-12y-l2x-\-(-4y-{-lx-6y}-6x-9y)-8x-\-y^}. 

18.  Enclose  the  last  three  terms  of  a  ■}- b  —  c  +  d  —  e  in  parentheses 
preceded  by  a  —  sign,  and  in  the  result  enclose  the  last  two  terms  in 
parentheses  in  brackets  preceded  by  a  —  sign. 


32       ADVANCED  COURSE  IN  ALGEBRA 


IV.    MULTIPLICATION  OF  ALGEBRAIC 
EXPRESSIONS 

83.  The  Rule  of  Signs. 

The  results  (8),  (9),  and  (12),  §  40,  hold  when  a  and  b  are 
any  positive  numbers. 

From  these  results  we  may  state  what  is  called  the  Rule  of 
Signs  in  multiplication,  as  follows  : 

The  product  of  two  terms  of  like  sign  is  positive;  the  product 
of  two  terms  of  unlike  sign  is  negative. 

84.  We  have  by  §  40,  (12), 

(—  a)  X  (—  6)  X  (—  c)  =  (ab)  X  (—  c) 

=  —  abc ;  (1) 

(-  a)  X  (-  6)  X  (-  c)  x{-d)  =  (-  abc)  x  (-  d),  by  (1), 

=  abcd;  etc. 

That  is,  the  product  of  three  negative  terms  is  negative ;  the 
product  of  four  negative  terms  is  positive ;  and  so  on. 

In  general,  the  product  of  any  number  of  terms  is  positive  or 
negative  according  as  the  number  of  negative  terms  is  even  or  odd. 

85.  The  Index  Law. 

Let  it  be  required  to  multiply  a^  by  a\ 
By  §  60,  a^  =a  xaxa, 

and  a^  =  ax  a. 

Whence,  a^xa^=axaxaxaxa  —  a^. 

We  will  now  consider  the  general  case. 

Let  it  be  required  to  multiply  a™  by  a'\  where  m  and  n  are 
any  positive  integers. 

We  have        a™  =  a  x  a  x  •  •  •  to  m  factors, 
and,  a"  =  a  X  a  X  •  •  •  to  n  factors. 

Then,     a'"  x  c^"  =  a  X  a  X  ••'  to  m-\-n  factors  =  a*""^**- 


MULTIPLICATION  33 

Hence,  the  exponent  of  a  letter  in  the  product  is  equal  to  its 
exponent  in  the  multiplicand  plus  its  exponent  in  the  multiplier. 

This  is  called  the  Index  Law  for  Multiplication. 

A  similar  result  holds  for  the  product  of  three  or  more 
powers  of  the  same  letter. 

Thus,  a^xa^xa^  =  a^+^+'  =  a^\ 

86.   Multiplication  of  Monomials. 

1.  Let  it  be  required  to  multiply  7  a  by  —  2  &. 
By  §72,  _2&  =  (-2)x6. 
Then,              7a  X  (-26)  =  7ax  (-2)  x  h. 

Then  by  the  Commutative  Law  for  Multiplication  (§  14), 
7ax  (-2  6)  =  7  X  (-2)  xax  6=  -14a&  (§83). 

2.  Required  the  product  of  —  2  a^h^,  6  ab^,  and  —  7  a^c. 
{-2a^W)X  6ab'x(-7a*c) 

=  (-  2)  a'b^  x6ab'x  (-  7)  a*c 

=  (-2)x6x(-7)xa^Xaxb^xb'xc 

=  84a36%  by  §§84and85. 

We  then  have  the  following  rule  for  the  product  of  any 
number  of  monomials : 

To  the  product  of  the  mcmerical  coefficients  (§§  72,  84,  85) 
annex  the  letters;  giving  to  each  an  exponent  equal  to  the  sum 
of  its  exponents  in  the  factors. 

3.  Multiply  -  5  a^ft  by  -  8  ah\ 

(_  5  a36)  X  (-  8  ab^)  =  40  a^^^b^"^^  =  40  a^b\ 

4.  Find  the  product  of  4  n^,  —  3  n^,  and  2  n\ 

4^2  X  (-  3n'')  X  2n^  =  -  24ri2+«+'«  =  -  24 n". 

5.  Multiply  —  flj"*  by  7  a;^,  m  being  a  positive  integer. 

(—  a;"')  X  7  a;^  =  —  7  of-^^ 

6.  Multiply  Q(m-[-  n)<  by  7  (m  +  nf. 

6  {m  +  n)*  X  7  (m  +  nf  =  42  (m  +  n)\. 


34       ADVANCED  COURSE  IN  ALGEBRA 

87.  Multiplication  of  Polynomials  by  Monomials. 

By  §§  14,  III,  and  24,  (7),  we  have  the  following  rule  for  the 
product  of  a  polynomial  by  a  monomial : 

Multiply  each  term  of  the  multiplicand  by  the  multiplier,  and 
add  the  partial  products. 

Ex.   Multiply  2x^-5x-\-7hj  -Sa^. 

{2a^-5x-^7)  X  (-8a^) 

=  (2aj2)x  (-8a^)  +  (-5aj)x  (-Sa?)  +  (7)  x  (-8a^) 

=  _16a^  +  40a;^-56aj3. 

The  student  should  endeavor  to  put  down  the  final  result  in  one 
operation. 

88.  Multiplication  of  Polynomials  by  Polynomials. 

By  the  Distributive  Law  for  Multiplication  (§  14), 

(a  +  6)  X  (c  +  d)  =  (a  +  6)  X  c  +  (a  +  6)  X  c? 

=  ac  -{-  be  +  ad  -\-bd ', 

and  a  similar  result  holds  whatever  the  number  of  terms  in 
the  multiplicand  or  multiplier. 

We  then  have  the  following  rule : 

Multiply  each  term  of  the  multiplicand  by  each  term  of  the 
multiplier,  and  add  the  partial  products. 

1.  Multiply  3a-4&  by  2a-5&. 

In  accordance  with  the  rule,  we  multiply  3  a  —  4  &  by  2  a, 
and  then  by  —5  b,  and  add  the  partial  products. 

A  convenient  arrangement  of  the  work  is  shown  below, 
similar  terms  being  in  the  same  vertical  column. 

3a  -46 
2a  -5b 
6a'-    Sab 

-15ab  +  20  b^ 
6  a:'-  23  ab  +  20  b\ 


MULTIPLICATION  35 

The  work  may  be  verified  by  performing  the  example  with  the  multi- 
plicand and  multiplier  interchanged. 

2.   Multiply  4  aa^  +  a^  -  8  a^  -  2  a^a;  by  2  a;  +  a. 

It  is  convenient  to  arrange  the  multiplicand  and  multiplier 
in  the  same  order  of  powers  of  some  common  letter  (§  76),  and 
write  the  partial  products  in  the  same  order. 

Arranging  the  expressions  according  to  the  descending  powers 

of  a.  we  have 

a3  _  2  a^a;  +  4  ay?  -  8  a;^ 

a  +2x 

a*  —  2  a^a;  +  4  a  V  —  8  ax^ 

2  a^aj  -  4  aV  +  8  aar^  -  16  aj^ 


a^  - 16  x\ 

If  the  multiplicand  and  multiplier  are  arranged  in  order  of 
powers  of  a  certain  letter,  with  literal  coefficients,  the  operation 
may  sometimes  be  abridged  by  the  use  of  parentheses. 

3.    Multiply  aj2  —  ax  ~hx  —  ab  by  x  —  a. 

By  §  87,  —  ax  —  hx  can  be  written  —  (a  +  h)x. 

x^  —      (a  +  l))x  —  ah 


y?  —     (a  +  V)x^  —  ahx 

—  ax^  4-  (<i^  +  ab)x  -\-  a^h 


a^  -  (2  a  +  6)a^  +  a'^  +  o?b, 

4.    Multiply  a;  —  m  by  a;  +  n. 


X  —m 

X  -\-n 

x"- 

mx 

+ 

nx  —  mn 

a^  -h  (—  m  -h  n)x  —  mn. 

It  is  convenient  to  write  the  coefficient  Qf  x  in  parentheses, 
when  adding  the  terms  —  mx  and  nx. 


36  ADVANCED  COURSE  IN   ALGEBRA 

89.  Homogeneity. 

If  the  multiplicand  and  multiplier  are  homogeneous  (§  Q>5), 
the  product  will  also  be  homogeneous,  and  its  degree  equal  to 
the  sum  of  the  degrees  of  the  multiplicand  and  multiplier. 

For  if  each  term  of  the  multiplicand  is  of  the  mth  degree 
(§  64),  and  each  term  of  the  multiplier  of  the  nth  degree,  each 
term  of  the  product  will  be  of  the  (m  +  n)th  degree  (§  85). 

The  examples  in  §  88  are  instances  of  the  above  law  ;  thus  in  Ex.  2, 
the  multiplicand,  multiplier,  and  product  are  homogeneous,  and  of  the 
third,  first,  and  fourth  degrees,  respectively. 

The  student  should  always,  when  possible,  apply  the  prin- 
ciples of  homogeneity  to  test  the  accuracy  of  algebraic  work. 

Thus,  if  two  homogeneous  expressions  be  multiplied  together, 
and  the  product  obtained  is  not  homogeneous,  it  is  evident  that 
the  work  is  not  correct. 

90.  Multiplication  by  Detached  Coefficients. 

In  finding  the  product  of  two  expressions  which  are  arranged 
according  to  the  same  order  of  powers  of  some  common  letter, 
the  operation  may  be  abridged  by  writing  only  the  numerical 
coefficients  of  the  terms. 

1.  Multiply  3x24-5a;-4by2a;2-7a;  +  l. 

3+5-4 
2-   7+    1 
6  +  10-    8 
_  21  -  35  +  28 

3+    5-4 

6-11-40  +  33-4. 

We  know  that  the  exponent  of  x  in  the  first  term  is  4. 
Then,  the  product  is  6  a:*  -  11  a^  -  40  a;^  +  33  a;  -  4. 
If  the  term  involving  any  power  be  wanting,  it  may  be 
supplied  with  the  coefficient  0. 

2.  Multiply  4  a^  +  6  aa;2  -  7  a:^  by  2  a^  -  3  a^l 

In  this  case  the  term  involving  a^x  in  the  multiplicand  and 
the  term  involving  ax  in  the  multiplier  are  wanting. 


MULTIPLICATION  37 

4  +  0+    6-    7 
2  +  0-    3 


8  +  0  +  12-14 

_  12  +  .  0  -  18  +  21 
8  +  0+    0-14-18  +  21. 

We  know  that  the  product  is  homogeneous  (§  89),  and  that 
the  exponent  of  a  in  the  first  term  is  5. 

Then,  the  product  is  8  a^  -  14  aV  -  18  ax^  +  21  x^. 

3.   Find  the  value  of  (2  a;  -  3)  (3  a;  +  5)  (6  a;  - 1). 

2-3 

3+5 


6 

-   9 

+  10-15 

6 
6 

+   1-15 
-    1 

36 

+   6-90 
-6-1 

+  15 

Kesult,  36  ar^  -91aj+15. 

91.   By  §  83, 

(+  ci)  X  (+  6)  =  +  ab,  (+  a)  X  (-  6)  =  -  ab, 

(-  a)  X  (-  6)  =  +  a6,  (-  a)  x  (+  6)  =  -  ab. 

Hence,  in  the  indicated  product  of  two  monomial  expressions, 
the  signs  of  both  expressions  may  be  changed  without  altering  the 
product;  but  if  the  sign  of  either  one  be  changed,  the  sign  of  the 
product  will  be  changed. 

The  above  is  true  for  the  product  of  a  monomial  and  a  poly- 
nomial, or  of  two  polynomials. 

If  either  expression  is  a  polynomial,  care  must  be  taken,  on 
changing  its  sign,  to  change  the  sign  of  each  of  its  terms. 

For  by  §  81,   -  (a  -  6  +  c)  =  -  o  +  6  -  c. 

Thus,  (a  —  b)(c  —  d)  may  be  written  in  the  forms 

(6  —  a)(d  —  c),   —(b  —  a)(c  —  d),  or  —  (a  —  6)  (d  —  c). 


38  ADVANCED  COURSE   IN  ALGEBRA 

In  like  manner  it  may  be  shown  that,  in  the  indicated  product 
of  more  than  two  expressions,  the  signs  of  any  even  number  of 
them  may  he  changed  without  altering  the  product;  but  if  the  signs 
of  any  odd  number  of  them  be  changed,  the  sign  of  the  product  will 
be  changed  (§  84). 

Thus,  (a  —  b)(c  —  d)  (e  — /)  may  be  written  in  the  forms 

(a-b)(d-c){f-e), 

(b-a){c-d)(f-e), 

—  (6  —  a)  (d  —  c)  (/—  e),  etc. 

EXERCISE  2 

Multiply  the  following : 

1.  x3  -  6  ic2  +  12  x  -  8  and  a:2  +  4  a;  +  4. 

2.  w  -  5  ^2  +  2  +  w3  and  5  n  +  w^  -  10. 

3.  3(a  +  6)2  _  (a  +  &)  +  2  and  4(a  +  6)2  _  («  +  6)  _  5. 

4.  4  x^wi+Si/M-s  4-  5  ajw+SySH-z  and  3  x^"^~^y^  —  7  x^y^^+K 

5.  x2  —  (m  —  n)  ic  —  wiw  and  a;  —  p. 

6.  x^  +  ax  -bx  —  ah  and  a;  +  6. 

7.  a^  +  3  +  5  c^4  _  6  a  _  2  a2  and  6  +  2  a2  -  a. 

8.  ma;  +  m?/  —  nx  —  ny  and  wa;  —  my  ■\-  nx  —  ny. 

9.  2  X  -  3  y,  3  a;  +  2  y,  2  X  +  3  y,  and  3  X  -  2  y. 

10.  X  +  a,  X  +  6,  and  x  —  c. 

11.  2     2_3  ;^  g^^^  lwi2  +  lwi-i 
3           4  2  3         9 

12.  x2  -  (a  +  6)  X  +  a6  and  x2  -  (c  —  d)  x  —  cd 

13.  a  +  6  +  c,  a  —  6  +  c,  and  a  +  6  —  c. 

Simplify  the  following : 

14.  [3x-(5?/  +  20)][3x-(5?/-2;2)]. 

15.  [(m  +  2n)-(2w- w)][(2wi  +  n)-(m-2w)]. 

16.  [2  x2  +  (3x  -  l)(4x  +  5)]  [5x2  -(4x  +  S)(x  -  2)]. 

17.  (a  -  6)  (a8  +  &»)  [a  (a  +  6)  +  ft^]. 


DIVISION  -  39 


V.    DIVISION  OP    ALGEBRAIC  EXPRESSIONS 

In  the  present  chapter,  we  shall  consider  those  cases  only  in  which  the 
Dividend,  Divisor,  and  Quotient  are  rational  and  integral  (§  63). 

In  such  cases,  the  division  is  said  to  be  exacts  and  the  dividend  is  said 
to  be  divisible  by  the  divisor. 

92.  The  Reciprocal  of  a  number  is  1  divided  by  that  number. 

Thus,  the  reciprocal  of  a  is  -• 

a 

93.  We  have  a  x  1  =  a. 

Regarding  a  as  the  quotient,  1  as  the  divisor,  and  a  as  the 
dividend,  we  have 

l  =  a.     . 

94.  By  §  30,  (1),       ^  =  ^xj  =  axi(§  93). 

0        1        0  0 

Hence,  to  divide  by  a  number  is  the  same  thing  as  to  multiply 
by  its  reciprocal  (§  92). 

95.  The  Commutative  Law  for  Division. 

By  §  94,  every  operation  in  Division  can  be  expressed  as  an 
operation  in  Multiplication. 

Thus,  if  a  is  to  be  divided  by  b,  c,  •••,  in  succession,  the 
result  is  ^      ^ 

ax-X-X— . 
b      c 

It  follows  from  this,  by  §  14,  I,  that  if  a  number  is  to  be 
divided  by  any  number  of  numbers  in  succession,  the  divisions 
can  be  performed  in  any  order. 

This  is  the  Commutative  Law  for  Division. 

It  may  be  expressed  as  follows : 

(a-j-6)  -!-C'"  =  (a-i-c)-^b-",  etc. 


96.  By  §  14,  I,  II,       b(ac)  =  a(bc). 


40  ADVANCED  COURSE  IN  ALGEBRA 

Then  by  §31,  ^  =  ^.  (1) 

That  is,  a  factor  common  to  the  dividend  and  divisor  can  he 
removed,  or  cancelled. 
Putting  6  =  1,  in  (1),  we  have 

"B^^^a  (§93). 

C  JL 

That  is,  if  a  number  he  hotli  multijMed  and  divided  hy  another, 
the  value  of  the  former  will  not  be  changed. 

97.  The  Rule  of  Signs. 

From  the  results  (1),  (2),  and  (3),  §  48,  we  may  state  the 
Rule  of  Signs  in  Division,  as  follows : 

TJie  quotient  of  two  terms  of  like  sign  is  positive;  the  quotient 
of  two  terms  of  unlike  sign  is  negative. 

98.  The  Index  Law  for  Division. 

Let  it  be  required  to  divide  a^  by  a^. 

By  §  60, 

a"  a  X  a 

Cancelling  the  common  factor  axa  (^  96),  we  have 

a^ 

We  will  now  consider  the  general  case : 
Let  it  be  required  to  divide  a"*  by  a**,  where  m  and  n  are  any 
positive  integers  such  that  m  is  >  n. 

We  have  —  =  a  X^  X  a  X  •♦•  to  m  factors 

'  a"      ax  ax  ax  •••tow  factors 

Cancelling  the  common  factor  axaxax--- to  n  factors, 

~  =  a  X  axax  •••torn  —  n  factors 
a" 

=  a**"". 

Then,  the  exponent  of  a  letter  in  the  quotient  is  equal  to  its 
exponent  in  the  dividend,  minus  its  exponent  in  the  divisor. 
This  is  called  the  Index  Law  for  Division. 


DIVISION  41 

99.  Division  of  Monomials. 

1.  Let  it  be  required  to  divide  —  14  a^h  by  7  a^. 
Bv  §  86  —  14  a^h  _  (—2)x7xa^xb 

^         '  7  a'  1X0?  ' 

Cancelling  the  common  factors  7  and  a?,  we  have 

z:ii^  =  (-2)x6=-26. 

We  then  have  the  following  rule  for  the  quotient  of  two 
monomials : 

To  the  quotient  of  the  numerical  coefficients  annex  the  letters, 
giving  to  each  an  exponent  equal  to  its  exponent  in  the  dividend 
minus  its  exponent  in  the  divisor,  and  omitting  any  letter  having 
the  same  exponent  in  the  dividend  and  divisor. 

2.  Divide  54  a^6V  by  -  9  a'h\ 

54  a'hh'' 
-^a'h^ 

3.  Divide  —  2  x^'^y'^z''  by  —  x'^y'^z^ ;  m,  n,  and  r  being  positive 
integers,  and  r  >  5. 

—  x^'y^'z^ 

4.  Divide  35  (a  -  by  by  7  (a  -  b)*. 

l{a-by         ^  ^ 

100.  Division  of  Polynomials  by  Monomials. 
By  §  94, 

a  a 

=  5  .-  +  c  .-  -  (^  .-,  by  §§  14,  III,  and  24,  (7), 
a  a  a 

=  -  +  ---,  by  §94. 
a      a      a 

Hence,  to  divide  a.  polynomial  by  a  monomial,  we  divide  each 

term  of  the  dividend  by  the  divisor,  and  add  the  results. 

This  is  called  the  DistributiveLgwforJDivision. 


42  ADVANCED  COURSE   IN   ALGEBRA 

1.  Divide  9  c^W  -^o^c-^  12  a%(?  by  -  3  al 

=  -3b'  +  2a'c-4.ahc\ 
The  student  should  endeavor  to  put  down  the  result  in  one  operation. 

2.  Divide  35  (x  +  yY  -  20  («  +  2/)^  by  5  ( a;  +  yf. 

35(x-{-yy-20(x  +  yy      ^  /     ,     -,2      ^/     .     n 

101.   Division  of  Polynomials  by  Polynomials. 
Ex.     Let  it  be  required  to  divide  12  + 10  or'  — 11  x  —  21  a^  by 
2a^-4.-3x. 

Arranging  the  expressions  according  to  the  descending  powers 
of  X  (§  76),  we  are  to  find  an  expression  which,  when  rnultiplied 
by  2rc2  -  3a;  -  4,  will  produce  10 a^  -  21  a^  -  11  a;  +  12. 

It  is  evident  that  the  term  containing  the  highest  power  of 
X  in  the  product  is  the  product  of  the  terms  containing  the 
highest  powers  of  x  in  the  multiplicand  and  multiplier. 

Therefore,  10  x^  is  the  product  of  2  a:^  and  the  term  containing 
the  highest  power  of  x  in  the  quotient. 

Whence,  the  term  containing  the  highest  power  of  x  in  the 
quotient  is  10  a^  divided  by  2  x^,  or  5  x. 

Multiplying  the  divisor  by  5x,  we  have  the  product  10  af^  — 
15  a^  —  20  a; ;  which,  when  subtracted  from  the  dividend,  leaves 
the  remainder  —  6  a;^  +  9  a;  +  12. 

This  remainder  must  be  the  product  of  the  divisor  by  the 
rest  of  the  quotient ;  therefore,  to  obtain  the  next  term  of  the 
quotient,  we  regard  —  6  a;^  +  9  a;  +  12  as  a  new  dividend. 

Dividing  the  term  containing  the  highest  power  of  a;,  —  6  a^, 
by  the  term  containing  the  highest  power  of  x  in  the  divisor, 
2  x^,  we  obtain  —  3  as  the  second  term  of  the  quotient. 

Multiplying  the  divisor  by  —  3,  we  have  the  product  —601^ 
+  9  a;  +  12  ;  which,  when  subtracted  from  the  second  dividend, 
leaves  no  remainder. 

Hence,  5  x  —  3  is  the  required  quotient. 


DIVISION  48 


It  is  customary  to  arrange  the  work  as  follows : 


10a^-15x'-20x 


2  ic^  —  3  a;  —  4,  Divisor. 


5  ic  —  3,  Quotient. 


-  6^2+    9a; +  12 

-  6a^+    9  a; +  12 


The  example  might  have  been  solved  by  arranging  the  dividend  and 
divisor  according  to  ascending  powers  of  x. 

102.   From  §  101,  we  derive  the  following  rule  : 

Arrange  the  dividend  and  divisor  in  the  same  order  of  powers 
of  some  common  letter. 

Divide  the  first  term  of  the  dividend  by  the  first  term  of  the 
divisor  J  and  write  the  result  as  the  first  term  of  the  quotient. 

Multiply  the  whole  divisor  by  the  first  term  of  the  quotient,  and 
subtract  the  product  from  the  dividend. 

If  there  be  a  remainder,  regard  it  as  a  new  dividend,  and 
proceed  as  before;  arranging  the  remainder  in  the  same  order  of 
powers  as  the  dividend  and  divisor. 

1.   Divide  9a62  +  a3-963-5a26 by  3&2-fci2_2a6. 

Arranging  according  to  the  descending  powers  of  a, 


a^-5a'b-{-9  ab'^  -9  b' 

a'-2ab-\-3b' 

a'-2a'b-\-3  ab'^ 
-Sa'b-\-6ab' 
-Sa'b-^6ab'-9b' 

a-3b 

In  the  above  example,  the  last  term  of  the  second  dividend  is  omitted, 
as  it  is  merely  a  repetition  of  the  term  directly  above. 

The  work  may  be  verified  by  multiplying  the  quotient  by  the  divisor, 
which  should  of  course  give  the  dividend. 

2.   Divide  4  +  9 a;^-28a;2  by  _3a^  +  2  +  4a;. 
Arranging  according  to  the  ascending  powers  of  x, 


4_28a;2+    9a;^ 
44    8a;  -    6a^ 


2  +  4a;-3a;2 


2_4aj-3a:2 


8a;  -22a;2_^    9^4 
8a;  -16a;2  +  12a;3 

-  6a;2-12ar^  +  9a;^ 

-  6x'-12o^i-9x* 


44       ADVANCED  COURSE  IN  ALGEBRA 

3.   Divide  x^ -\- {a  +  b  —  c)  a^ -{-  (ab  —  be  —  cd)  x  —  abe  by  a;  +  a. 


a^  -{-  (a  +  b  —  e)x^  +  (ab  - 

-be 

—  ca)x  - 

-  abe 

X  -\-a 

ar^+                  ax" 

x^  +  {b  —  e)x—bc 

{b-e)x' 

(b-e)x^-{-(ab 

—  ca)x 

—  bex 

—    bex 

—  abc 

103.  It  is  evident  from  §  89  that,  if  the  dividend  and  divisor 
are  homogeneous,  the  quotient  will  also  be  homogeneous,  and  its 
degree  equal  to  the  degree  of  the  dividend  minus  the  degree  of 
the  divisor. 

104.  Division  by  Detached  Coefficients. 

In  finding  the  quotient  of  two  expressions  which  are  arranged 
according  to  the  same  order  of  powers  of  some  common  letter, 
the  operation  may  be  abridged  by  writing  only  the  numerical 
coefficients  of  the  terms. 

If  the  term  involving  any  power  is  wanting,  it  may  be  sup- 
plied with  the  coefficient  0. 

Ex.   Divide  ea^-\-2a^-9x*-\-5x'-i-lSx~80hy3a^+x^-6. 

3+1+0-6 


6  +  2-9+   0  +  5  +  18- 
6  +  2  +  0-12 

-9  +  12 

_9-   3  +  0  +  18 

-30 

15  +  5 

15  +  5+   0- 

-30 

2+0-3+5 


Then  the  quotient  is  2  a^  —  3  a;  +  5. 

105.   By  §  37,  (6),  if  a  is  any  number, 
0  X  a  =  0. 

Regarding  0  as  the  quotient,  a  as  the  divisor,  and  0  as  the 

dividend,  we  have  a 

-  =  0. 
a 


DIVISION  45 

EXERCISE  3 

Divide  the  following : 

1.  x5  +  37  x2  -  70  ic  +  50  by  a;2  -  2  a;  +  10. 

2.  6(x~yy-7(x-y)-20  hj  S{x-y)  +  i. 

3.  a^+^b^  +  a&39+2  i^y  Qtp+152  ^  ab^+^ ;  p  and  ^  being  positive  integers. 

4.  a&  -  &5  _  5  a^b  ^^ab^  +  10  a^S^  -  10  a^b^  hy  a^-b^-S  a'^b  +  3  ab^. 

5.  6  n^  +  25  #  -  7  n3  _  81 7i2  _  3  ri  +  28  by  2  n^  +  5  n2  -  8  n  -  7. 

6.  23x2  -  5  x4  -  12  +  12  a;5  +  8  X  -  14  a:3  by  x  -  2  +  3  x2. 

7.  16(a  +  ft)4  -  81  by  2(«  +  6)  -  3. 

8.  8  x6  -  4  x5  -  2  X*  +  15  x3  +  3  x2  -  5  X  -  15  by  4  x»  -  a;  -  3. 

9.  a^  +  b^  -  c^  +  S  abc  hy  a  +  b  -  c. 

10.  im4-2m«  +  -m2--L  by  ?m2-^wi--. 
9  4  16     "^  3  2  4 

11.  52  x3  +  64  +  18  X*  -  200  x2  -f  x5  by  6  x2  -  8  +  x^  -  12  x. 

12.  a^-6  o4,j2  ^  9  Qj2„4  _  4  ^6  by  cfS  _  2  a2^  _  an2  +  2  #. 

13.  x^  -  (a  —  6  +  c)x2  —  (ab  ~  ac -^  bc)x  +  a6c  by  x^  —  (a  —  b)x  —  a6. 

14.  x3  +  (a  -  6  4-  c)x2  +  (_  a6  +  ac  -  &c)x  —  abc  by  x  +  c. 

15.  x3  4-  (4  a  +  2  6  4-  3  c)x2  4-  (8  a6  +  12  ac  +  6  bc)x  4-  24  abc  by 
x2  +  (4  a  4-  3  c)x  4- 12  ac. 

16.  x3  -  (a  +  3  &  +  2  c)x2  4-  (3  a6  4-  2  ac  4-  6  bc)x  ~  6  abc  hy  x-Sb. 

17.  (2  w2  4- 10  mn)  x^ -\- (S  m^  -  9  mn  -  15  w2)x  -  (12  mn-9  n^)    by 
2  mx  —  3  n. 

18.  x*-(4a4-3)x3  +  (12a-5&4-2)x2-(8a-15  6)x- 106    by 
x2  -  3  X  4-  2. 


46  ADVANCED  COURSE  IN  ALGEBRA 


VI.    INTEGRAL    LINEAR   EQUATIONS 

106.  The  First  Member  of  an  equation  is  the  expression  to 
the  left  of  the  sign  of  equality,  and  the  Seco7id  Member  the 
expression  to  the  right  of  that  sign. 

Thus,  in  2  a?  —  3  =  3  a;  +  5,  the  first  member  is  2  oj  —  3  and 
the  second  member  3  x  +  5. 

Any  term  of  either  member  of  an  equation  is  called  a  term 
of  the  equation. 

The  sides  of  an  equation  are  its  two  members. 

107.  An  Identical  Equation,  or  Identity,  is  an  equation  whose 
members  are  the  same,  or  become  the  same  after  all  the  indi- 
cated operations  have  been  performed ;  as, 

5  =  5,  or  (a  -{-  b)  (a  —  b)  —  a?  —  W. 

The  sign  =,  read  "is  identically  equal  to,''''  is  frequently  usedrin  place 
of  the  sign  of  equality  in  an  identity. 

It  is  evident  that,  in  an  identity  involving  letters,  the  mem- 
bers are  equal  whatever  values  are  given  to  the  letters,  provided 
the  same  value  is  given  to  the  same  letter  wherever  it  occurs. 

All  equations  considered  up  to  the  present  time  have  been 
identical  equations. 

108.  An  equation  is  said  to  be  satisfied  by  a  set  of  values  of 
certain  letters  involved  in  it  when,  on  substituting  the  value  of 
each  letter  in  place  of  the  letter  wherever  it  occurs,  the  equa- 
tion becomes  identical. 

Thus,  the  equation  x  —  y  =  5  is  satisfied  by  the  set  of  values 
x  =  S,  y  =  S;  for,  on  substituting  8  for  x,  and  3  for  y,  the  equa- 
tion becomes 

8  —  3  =  5,  or  5  =  5;  which  is  identical. 

109.  An  Equation  of  Condition  is  an  equation  involving  one 
or  more  letters,  called  Unknown  Numbers,  which  is  satisfied 
only  by  particular  values,  or  sets  of  values,  of  these  letters. 


INTEGRAL  LINEAR  EQUATIONS  47 

Thus,  the  equation  x-{-2  =  5  is  satisfied  by  the  value  x  =  3; 
but  not  by  the  value  x  =  5. 

Again,  the  equation  x-{-y  =  7  is  satisfied  by  the  set  of  values 
a;  =  4,  2/  =  3 ;  but  not  by  the  set  of  values  x  =  6,  y  =  9. 

An  equation  of  condition  is  usually  called  an  equation. 

Any  letter  in  an  equation  of  condition  may  represent  an 
unknown  number ;  but  it  is  usual  to  represent  unknown  num- 
bers by  the  last  letters  of  the  alphabet. 

110.  Any  letter,  or  set  of  letters,  which  satisfies  an  equation 
is  called  a  Solution  of  the  equation. 

If  the  equation  contains  but  one  unknown  number,  its  solu- 
tions are  called  Roots. 

A  solution  is  verified  when,  on  substituting  the  values  of  the 
unknown  numbers,  and  performing  the  operations  indicated, 
the  equation  becomes  identical. 

To  solve  an  equation,  or  a  system  of  equations,  is  to  find  its 
solutions. 

111.  A  Numerical  Equation  is  one  in  which  all  the  known 
numbers  are  represented  by  Arabic  numerals ;  as, 

2x-7  =  x  +  6. 

A  Literal  Equation  is  one  in  which  some  or  all  of  the  known 
numbers  are  represented  by  letters ;  as, 

Sx-\-a  =  5x  —  2b. 

An  Integral  Equation  is  one  each  of  whose  members  is  a 
rational  and  integral  expression  (§  63);  as, 

4ic  — 5  =  -v  +  l. 

112.  If  a  rational  and  integral  monomial  (§  63)  involves  a 
certain  letter,  its  degree  with  respect  to  it  is  denoted  by  its 
exponent. 

If  it  involves  two  letters,  its  degree  with  respect  to  them  is 
denoted  by  the  sum  of  their  exponents ;  etc. 

Thus,  2  ab^x^y^  is  of  the  second  degree  with  respect  to  cc,  and 
of  the  fifth  with  respect  to  x  and  y. 


48  ADVANCED   COURSE  IN   ALGEBRA 

113.  If  an  integral  equation  (§  111)  contains  one  or  more 
unknown  numbers,  the  degree  of  the  equation  is  the  degree  of 
its  term  of  highest  degree. 

Thus,  if  X  and  y  represent  unknown  numbers, 

ax  —  hy=  c,  is  an  equation  of  t\iQ  first  degree ; 
x^  -\-4,x=  —  2,  an  equation  of  the  second  degree ; 
2  ic"  —  3  xf  =  5,  an  equation  of  the  third  degree ;  etc. 

A  Linear  Equation  is  an  equation  of  the  first  degree. 

114.  Two  equations,  each  involving  one  or  more  unknown 
numbers,  are  said  to  be  Equivalent  when  every  solution  of  the 
first  is  a  solution  of  the  second,  and  every  solution  of  the 
second  a  solution  of  the  first. 

116.   The  following  are  of  use  in  solving  equations : 

1.  If  the  same  number  (or  equal  numbers)  be  added  to  equal 
^lumbers,  the  sums  will  be  equal. 

2.  If  the  same  number  (or  equal  numbers)  be  subtracted  from 
equal  numbers,  the  remainders  will  be  equal. 

These  follow  from  §  22. 

3.  If  equal  numbers  be  multiplied  by  the  same  number  (or 
equal  numbers),  the  products  will  be  equal  (§  29). 

4.  If  equal  numbers  be  divided  by  the  same  number  (or  equal 
numbers),  the  quotients  will  be  equal,  provided  the  divisor  is  not  0. 

This  follows  from  §  29 ;  compare  §  43. 

PRINCIPLES  USED  IN  SOLVING  INTEGRAL  EQUATIONS 

116.   Addition. 

If  the  same  expression  be  added  to  both  members  of  an  equation, 
the  resulting  equation  will  be  equivalent  to  the  first. 
Consider,  for  example,  the  equation 

A  =  B.  (1) 

To  prove  that  the  equation 

A^-C  =  B+G,  (2) 

where  C  is  any  expression,  is  equivalent  to  (1). 


INTEGRAL  LINEAR  EQUATIONS        49 

Any  solution  of  (1),  when  substituted  for  the  unknown  num- 
bers, makes  A  identically  equal  to  B  (§  108). 

It  then  makes  A-\-  G  identically  equal  to  J5  -f-  (7  (§  115,  1). 

Then  it  is  a  solution  of  (2). 

Again,  any  solution  of  (2),  when  substituted  for  the  unknown 
numbers,  makes  A-\-  C  identically  equal  to  B  -\-  G. 

It  then  makes  A  identically  equal  to  B  (§  115,  2). 

Then  it  is  a  solution  of  (1). 

Therefore,  (1)  and  (2)  are  equivalent. 

The  above  demonstration  proves  that  if  the  same  expression  be  sub- 
tracted from  both  members  of  an  equation,  the  resulting  equation  will  be 
equivalent  to  the  first. 

117.  Transposing  Terms. 

Consider  the  equation    x-\-a  —  b  =  c. 

Adding  —  a  and  +  &  to  both  members  (§  116),  we  have  the 
equivalent  equation 

x-{-a  —  b—a-\-b  =  c  —  a-\-b. 
Or,  x  =  c  —  a-\-b. 

In  this  case,  the  terms  a  and  —  b  are  said  to  be  transposed 
from  the  first  member  to  the  second. 

Hence,  if  any  term  be  transposed  from  one  member  of  an  equor 
Hon  to  the  other  by  changing  its  sign,  the  resulting  equation  will  be 
equivalent  to  the  first. 

If  the  same  term  appears  in  both  members  of  an  equation  affected  with 
the  same  sign,  it  may  be  cancelled. 

118.  Consider  the  equation 

a  —  x  =  b  —  c.  (1) 

Transposing  each  term,  we  have  the  equivalent  equation 

—  b-\-c=  — a-f-ic,  or  a;  —  a  =  c  —  ^; 

which  is  the  same  as  (1)  with  the  sign  of  every  term  changed. 

Hence,  if  the  signs  of  all  the  terms  of  an  equation  be  changed, 
the  resulting  equation  will  be  equivalent  to  the  first. 


50  ADVANCED   COURSE  IN  ALGEBRA 

119.  Multiplication. 

If  the  members  of  an  equation  he  multiplied  by  the  same  expres- 
sion, which  is  not  zero,  and  does  not  involve  the  unknown  numbers, 
the  resulting  equation  will  be  equivalent  to  the  first. 

Consider  the  equation       A=zB,  (1) 

To  prove  that  the  equation 

AxO=BxO,  (2) 

where  C  is  not  zero,  and  does  not  involve  the  unknown  num- 
bers, is  equivalent  to  (1). 

Any  solution  of  (1),  when  substituted  for  the  unknown  num- 
bers, makes  A  identically  equal  to  B. 

It  then  makes  AxC  identically  equal  to  5  x  O  (§  115,  3). 

Then  it  is  a  solution  of  (2). 

Again,  any  solution  of  (2),  when  substituted  for  the  unknown 
numbers,  makes  AxC  identically  equal  to  B  x  G. 

It  then  makes  A  identically  equal  to  B  (§  115,  4). 

Then  it  is  a  solution  of  (1). 

Therefore,  (1)  and  (2)  are  equivalent. 

The  reason  why  the  above  does  not  hold  for  the  multiplier  zero  is,  that 
the  principle  of  §  115,  4,  is  restricted  to  cases  where  the  divisor  is  not  zero. 

120.  The  necessity  for  limiting  the  principle  of  §  119  to 
cases  where  the  multiplier  does  not  involve  the  unknown  num- 
bers is  that,  if  C  contains  the  unknown  numbers,  the  equation 
AxO=BxO  is  satisfied  by  certain  values  of  the  unknown 
numbers  which  make  (7=0. 

But  these  values  do  not,  in  general,  satisfy  A=B. 
Consider,  for  example,  the  equation 

x-\-2  =  3x-4:.  (1) 

Now  the  equation 

(x  +  2)(x-l)  =  (Sx-4:)(x-l),  (2) 

which  is  obtained  from  (1)  by  multiplying  both  members  by 
a?  —  1,  is  satisfied  by  the  value  x  =  l,  which  does  not  satisfy  (1). 
Then  (1)  and  (2)  are  not  equivalent. 


INTEGRAL  LINEAR  EQUATIONS  51 

It  follows  from  this  that  it  is  never  allowable  to  multiply- 
both  members  of  an  integral  equation  by  an  expression  which 
involves  the  unknown  numbers  j  for  in  this  way  additional 
solutions  are  introduced. 

121.  Clearing  of  Fractions. 

Consider  the  equation 

2  5     5        9 

-X =  -X . 

3  4     6        8 

Multiplying  each  term  by  24,  the  lowest  common  multiple  of 
the  denominators  (§  119),  we  have  the  equivalent  equation 

16x-30  =  20a;-27, 

where  the  denominators  have  been  removed. 

Removing  the  fractions  from  an  equation  by  multiplication  is  called 
'•''Clearing  the  equation  of  fractions.'''' 

122.  Division. 

If  the  members  of  an  equation  he  divided  by  the  same  expres- 
siouj  which  is  not  zero,. and  does  not  involve  the  unknown  num- 
bers, the  resulting  equation  will  be  equivalent  to  the  first. 

Consider  the  equation      A  =  B.  (!) 

To  prove  that  the  equation 

A-^C=B-^C,  (2) 

where  C  is  not  zero,  and  does  not  involve  the  unknown  num- 
bers, is  equivalent  to  (1). 

Any  solution  of  (1),  when  substituted  for  the  unknown  num- 
bers, makes  A  identically  equal  to  B. 

It  then  makes  A-~-C  identically  equal  to  B-i-  C  (%  115,  4). 

Then  it  is  a  solution  of  (2). 

Again,  any  solution  of  (2),  when  substituted  for  tjie  unknown 
numbers,  makes  A-i-C  identically  equal  to  B -i- C. 

It  then  makes  A  identically  equal  to  B. 

Then  it  is  a  solution  of  (1). 

Therefore,  (1)  and  (2)  are  equivalent. 


52  ADVANCED   COURSE   IN   ALGEBRA 

123.  The  necessity  for  limiting  the  principle  of  §  122  to 
cases  where  the  divisor  does  not  involve  the  unknown  num- 
bers is  that,  if  C  contains  the  unknown  numbers,  the  solution 
oi  A  =  B  contains  certain  numbers  which  do  not,  in  general, 
satisfy  ^  ^  (7=  5 -^  a 

Consider,  for  example,  the  equation 

{x  +  2){x-l)  =  {3x-^){x-.l).  (1) 

Also  the  equation       a?  +  2  =  3  a;  —  4,  (2) 

which  is  obtained  from  (1)  by  dividing  both  members  by  x  —  1. 

Now  equation  (1)  is  satisfied  by  the  value  ic  =  1,  which  does 
not- satisfy  (2). 

Then  (1)  and  (2)  are  not  equivalent. 

It  follows  from  this  that  it  is  never  allowable  to  divide  both 
members  of  an  integral  equation  by  an  expression  which  in- 
volves the  unknown  numbers;  for  in  this  way  solutions  are 
lost. 

SOLUTION  OF  INTEGRAL  LINEAR  EQUATIONS 

124.  To  solve  an  equation  containing  one  unknown  number, 
we  put  it  into  a  succession  of  forms,  which  lead  finally  to  a 
knowledge  of  the  root  or  roots. 

This  process  is  called  transforming  the  equation. 

If  every  transformation  is  effected  by  means  of  the  principles 
of  §§  116  to  122,  the  successive  equations  will  have  the  same 
roots  as  the  given  equation,  and  no  solutions  will  be  introduced 
nor  lost. 

EXAMPLES 

125.  1.    Solve  the  equation 

5x-l  =  ?>x-\-l. 

Transposing  3  a;  to  the  first  member,  and  —  7  to  the  second 
(§117),wehave        5^_3^^7  +  l. 

Uniting  similar  terms,     2  a;  =  8. 


INTEGRAL   LINEAR  EQUATIONS  53 

Dividing  both  members  by  2  (§  122), 

X  =  A. 
To  verify  the  result,  put  x  =  4  in  the  given  equation. 
Thus,  20  -  7  =  12  +  1 ;  which  is  identical. 

2.  Solve  the  equation 

7         5_3         1 

Clearing  of  fractions  by  multiplying  each  term  by  60,  the 
L.  C.  M.  of  6,  3,  5,  and  4,  we  have 

70x-100  =  36a:-15. 

Transposing  36  x  to  the  first  member,  and  — 100  to  the 
second,  and  uniting  similar  terms, 

34  a;  =  85. 

Dividing  by  34,  ^  =  1  =  1 

3.  Solve  the  equation 

(5  -  3  ic)  (3  +  4  r»)  =  62  -  (7  -  3  a;)(l  -  4  x). 
Expanding,  15  +  11  ic-12  a^  =  62  -  (7  -  31  x  +  12  x") 

=  62  -  7  +  31  a;  -  12  a;2. 
Cancelling  the  —  12  x^  terms  (§  117),  and  transposing, 

11  a;  -  31  a;  =  62  -  7  -  15. 
Uniting  terms,  —  20  a;  =  40. 

Dividing  by  -  20,  x  =  -  2. 

To  expand  an  algebraic  expression  is  to  perform  the  operations  indicated. 

4.  Solve  the  equation 

.2  a;  -f  .001  -  .03  a;  =  .113  x  -  .0161. 
Transposing, 

.2  aj  -  .03  a;  -  .113  x  =  -  .0161  -  .001. 
Uniting  terms,  .057  x  =  —  .0171. 

Dividing  by  .057,  x  =  —  .3. 

From  the  above  examples,  we  have  the  following  rule  for 
solving  an  integral  linear  equation  with  one  ujiknown  number : 


54        ADVANCED  COURSE  IN  ALGEBRA 

Clear  the  equation  of  fractions,  if  any,  by  multiplying  each  term 
by  the  L.  C.  M.  of  the  denominators  of  the  fractional  coefficients. 

Remove  the  parentheses,  if  any. 

Transpose  the  unknown  terms  to  the  first  member,  and  the 
known  to  the  second;  cancelling  any  term  which  has  the  same 
coefficient  in  both  members. 

Unite  similar  terms,  and  divide  both  members  by  the  coefficient 
of  the  unknown  number. 

The  student  should  endeavor  to  apply  more  than  one  principle  at  one 
operation. 

He  will  also  find  it  excellent  practice  to  verify  his  solutions. 


«  ,       ,     .  „  EXERCISE  4 

Solve  the  following : 

5  3         9         15 

2.  .05  jc- 1.82- .7  a:  =  .008  a; -.504. 

3.  4(a;  +  14)  -  4(3  a:  -  32)  =  6(a;  +  12)  -  7  (x  -  12). 

.577  1  55 

4.  -X x  =  -  X X 

6  8         9         18         48 

5.  (5-3a;)(3 +4a;)-(7  +  3a;)(l-4a;)=-l. 

6.  .07(8  a:  -  5.7)  =  .8(5  x  +  .86)  +  1.321. 

7.  (1  +  3  xY  -  (5  -  xy  -  4(1  -  a;)  (3  -  2  X)  =  0. 

8.  6(a;-4)2  =  5-(3  -  2  x)2  -  5(2  +  a;)(7  -  2  a:). 

9.  (3  a;  -  2)3  -  9  a;  (a;  -  1)  (3  a;  -  8)  =  45  3^2  _  33. 

10.  (x  +  4)8-(a;-4)8  =  2(3a;-2)(4a;  +  l). 

11.  |(4  +  a:)-^(l-5x)  =  |(l  +  2a;)-A(2-3x). 

12.  |[.-l(5.  +  l)]  =  |[.-|(3.  +  4)]  +  I. 


PROBLEMS     INVOLVING    INTEGRAL    LINEAR    EQUATIONS 
WITH  ONE  UNKNOWN  NUMBER 

126.   For  the  solution  of  a  problem  by  algebraic  methods  no 
general  rule  can  be  given. 
The  following  suggestions  will  be  found  of  service  : 


INTEGRAL  LINEAR  EQUATIONS         55 

1.  Represent  the  unknown  number,  or  one  of  the  unknown 
numbers  if  there  are  several,  by  x. 

2.  Every  problem  contains,  explicitly  or  implicitly,  just  as 
many  distinct  statements  as  there  are  unknown  numbers  involved. 

Use  all  but  one  of  these  to  express  the  other  unknown  num- 
bers in  terms  of  x. 

3.  Use  the  remaining  statement  to  form  an  equation. 

ILLUSTRATIVE  PROBLEMS 

127.  1.  Divide  45  into  two  parts  such  that  the  less  part 
shall  be  one-fourth  the  greater. 

Here  there  are  two  unknown  numbers  ;  the  greater  part  and  the  less. 

In  accordance  with  the  first  suggestion  of  §  126,  we  represent  the 
greater  part  by  x. 

The  first  statement  of  the  problem  is,  implicitly : 

The  sum  of  the  greater  part  and  the  less  is  45. 

The  second  statement  is : 

The  less  part  is  one-fourth  the  greater. 

In  accordance  with  the  second  suggestion  of  §  126,  we  use  the  second 
statement  to  express  the  less  part  in  terms  of  x. 

Thus,  the  less  part  is  represented  by  -x. 

4 
We  now,  in  accordance  with  the  third  suggestion,  use  the  first  state- 
ment to  form  an  equation. 

Thus,  x-^-x  =  i6. 

4 

Clearing  of  fractions,  ix-{-x  =  180. 

Uniting  terms,  5x=  180. 

Dividing  by  5,  x  =  36,  the  greater  part. 

Then,  i  x  =  9,  the  less  part. 

4 

2.  A  is  3  times  as  old  as  B,  and  8  years  ago  he  was  7  times 
as  old.     Required  their  ages  at  present. 

Let  X  =  number  of  years  in  B's  age. 

Then,  3x  =  number  of  years  in  A's  age. 

Also,  X  —  S  =  number  of  years  in  B's  age  8  years  ago, 

and  3x  —  8  =  number  of  years  in  A's  age  8  years  ago. 


56       ADVANCED  COURSE  IN  ALGEBRA 

But  A's  age  8  years  ago  was  7  times  B's  age  8  years  ago. 

Whence,  3x-8  =  7(x-8). 

Expanding,  3x  —  8  =  7x—  56. 

Transposing,  —  4x=  —  48. 

Dividing  by  —  4,  x  =  12,  the  number  of  years  in  B's  age. 

Then,  3x  =  36,  the  number  of  years  in  A's  age. 

It  must  be  carefully  borne  in  mind  that  x  can  only  represent  an 
abstract  number. 

Thus,  in  Ex.  2,  we  do  not  say  "let  x  represent  B's  a^e,"  but  "let  x 
represent  the  number  of  years  in  B's  age." 

3.  A  sum  of  money,  amounting  to  f  4.32,  consists  of  108 
coins,  all  dimes  and  cents ;  how  many  are  there  of  each  kind  ? 

Let  X  =  the  number  of  dimes. 

Then,  108  —  x  =  the  number  of  cents. 

Also,  the  X  dimes  are  worth  10  x  cents. 
But  the  entire  sum  amounts  to  432  cents. 
Whence,  10  ic  +  108  -  x  =  432. 

Transposing,  9x  =  324. 

Whence,  x  —  36,  the  number  of  dimes ; 

and  108  —  x  =  72,  the  number  of  cents. 

4.  At  what  time  between  3  and  4  o'clock  are  the  hands  of 
a  watch  opposite  to  each  other? 

Let  X  =  the  number  of  minute-spaces  passed  over  by  the  minute-hand 
from  3  o'clock  to  the  required  time. 

Then  since  the  hour-hand  is  15  minute-spaces  in  advance  of  the  minute- 
hand  at  3  o'clock,  X  —  15  —  30,  or  x  —  45  will  represent  the  number  of 
minute-spaces  passed  over  by  the  hour-hand. 

But  the  minute-hand  moveS'  12  times  as  fast  as  the  hour-hand. 

Then,  x  =  12  (x  -  46) 

=  12  X  -  540. 
Transposing,  _  11  x  =  —  540. 

Whence,  x  =  49j\. 

Then,  the  required  time  is  i9j\  minutes  after  3  o'clock. 


INTEGRAL   LINEAR  EQUATIONS  57 

5.  Two  persons,  A  and  B,  63  miles  apart,  start  at  the  same 
time,  and  travel  towards  each  other.  A  travels  at  the  rate  of 
4  miles  an  hour,  and  B  at  the  rate  of  3  miles  an  hour.  How 
far  will  each  have  travelled  when  they  meet  ? 

Let  ^x  =  number  of  miles  that  A  travels. 

Then,  l(}^  =  number  of  miles  that  B  travels. 

By  the  conditions^  x-i-j^x  =  6^^ 

Then,  4  x  =  36,  number  of  miles  that  A  travels, 

and  3  a;  =  27,  number  of  miles  that  B  travels. 

It  is  often  advantageous,  as  in  Ex.  5,  to  represent  the  unknown  num- 
ber by  some  multiple  of  x,  instead  of  by  x  itself; 


EXERCISE  5 

1.  Divide  66  into  two  parts  such  that  -  the  greater  shall  exceed  - 
the  less  by  21.  '^  ^ 

5  S 

2.  In  9  years,  B  will  be  -  as  old  as  A  ;  and  12  years  ago  he  was  -  as 

old.     What  are  their  ages  ? 

(Let  x  represent  number  of  years  in  A's  age  12  years  ago.) 

3.  Divide  197  into  two  parts  such  that  the  smaller  shall  be  contained 
in  the  greater  5  times,  with  a  remainder  23. 

4.  After  A  has  travelled  7  hours  at  the  rate  of  10  miles  in  3  hours, 
B  sets  out  to  overtake  him,  travelling  at  the  rate  of  9  miles  in  2  hours. 
How  far  will  each  have  travelled  when  B  overtakes  A  ? 

5.  At  what  time  between  8  and  9  o'clock  are  the  hands  of  a  watch 
together  ? 

6.  Find  four  consecutive  odd  numbers  such  that  the  product  of  the 
first  and  third  shall  be  less  than  the  product  of  the  second  and  fourth 
by  86. 

7.  A  sum  of  money,  amounting  to  $19.30,  consists  of  .^2  bills,  25-cent 

pieces,  and  5-cent  pieces.    There  are  13  more  5-cent  pieces  than  $2  bills, 

7 
and  -  as  many  5-cent  pieces  as  25-cent  pieces.     How  many  are  there  of 

each  ? 

8.  At  what  times  between  4  and  5  o'clock  are  the  hands  of  a  watch  at 
right  angles  to  each  other  ? 


58  ADVANCED  COURSE   IN   ALGEBRA 

9.  A  woman  sells  half  an  egg  more  than  half  her  eggs.  She  then  sells 
half  an  egg  more  than  half  her  remaining  eggs.  A  third  time  she  does  the 
same,  and  now  has  3  eggs  left.     How  many  had  she  at  first  ? 

10.  A  train  leaves  A  for  B,  210  miles  distant,  travelling  at  the  rate  of 
28  miles  an  hour.  After  it  has  been  gone  1  hour  and  15  minutes,  another 
train  starts  from  B  for  A,  travelling  at  the  rate  of  22  miles  an  hour.  How 
many  miles  from  B  will  they  meet  ? 

11.  A  man  puts  a  certain  sum  in  a  savings  bank  paying  4%  interest. 
At  the  end  of  a  year  he  deposits  the  interest,  receiving  interest  on  the 
entire  amount.  At  the  end  of  a  second  year  and  a  third  year  he  does  the 
same,  and  now  has  $2812. 16  in  the  bank.     What  was  his  original  deposit  ? 

12.  A  fox  is  pursued  by  a  hound,  and  has  a  start  of  77  of  her  own 
leaps.  The  fox  makes  5  leaps  while  the  hound  makes  4  ;  but  the  hound 
in  5  leaps  goes  as  far  as  the  fox  in  9.  How  many  leaps  does  each  make 
before  the  hound  catches  the  fox  ? 

13.  A  clock  has  an  hour-hand,  a  minute-hand,  and  a  second-hand,  all 
turning  on  the  same  centre.  At  12  o'clock  alt  the  hands  point  at  12.  How 
many  seconds  will  it  be  before  the  hour-hand  is  between  the  other  two 
hands  and  equally  distant  from  them  ?  .         (^ 

14.  A  freight  train  travels  from  ^  to  ^  at  the  rate  of  12  miles  an  hour. 
After  it  has  been  gone  3^  hours,  an  express  train  leaves  A  for  B,  travel- 
ling at  the  rate  of  45  miles  an  hour,  and  reaches  B  1  hour  and  5  minutes 
ahead  of  the  freight.  Find  the  distance  from  Ato  B  and  the  time  taken 
by  the  express  train. 

15.  A  merchant  increases  his  capital  each  year  by  one-third,  and  at 
the  end  of  each  year  sets  aside  $  1350  for  expenses.    At  the  end  of  three 

145 
years,  after  setting  aside  his  expenses,  he  finds  that  he  has  — -  of  his 

original  capital.     What  was  his  original  capital  ? 


i 


SPECIAL  METHODS  59 


VII.  SPECIAL  METHODS  IN  MULTIPLICATION 
AND  DIVISION 

128.  Any  Power  of  a  Power. 

Required  the  value  of  (a"*)%  where  m  and  n  are  any  positive 
integers. 

We  have,  (a'")"=:  a'"  x  a"*  X  •••  to  n  factors  (§  60) 

Qm  +  TO  +  •  •  •  to  n  terms  ^_.  n^n 

129.  Any  Power  of  a  Product. 

Required  the  value  of  (abc ---y,  where  n  is  any  positive 
integer. 

We  have,  (abc-'-y 

=  {abc  •••)  X  (abc  •••)  x  •••  to  n  factors 

=  (ax  ax  '"ton  factors)  (b  xb  x  -"to  n  factors)  ••• 

130.  Any  Power  of  a  Monomial. 

1.  Required  the  value  of  (5  a-by. 

We  have,  (5  a'bf  =  5a'bx5a'bx5  a'b  =  125  a'b\ 

2.  Required  the  value  of  (—  my. 

We  have,  (—  my  =  (—  m)  x  (—  m)  x  (—  m)  x  (—  m)  =  m\ 

3.  Required  the  value  of  (—  3  n^y. 

We  have,  (- 3 n^y  =  (-3n')  x  {-Sn')  x  {-Sn')=  -27 n\ 

From  §§  128  and  129,  and  the  above  examples,  we  have  the 
following  rule  for  raising  a  rational  and  integral  monomial 
(§  63)  to  any  power  whose  exponent  is  a  positive  integer : 

Raise  the  absolute  value  of  the  numerical  coefficient  to  the 
required  power,  and  multiply  the  exponent  of  each  letter  by  the 
exponent  of  the  required  power. 

Give  to  every  poiver  of  a  positive  term,  and  to  every  even  power 
of  a  negative  term,  the  positive  sign  ;  and  to  every  odd  power  of 
a  negative  term  the  negative  sign. 


60       ADVANCED  COURSE  IN  ALGEBRA 

131.  Square  of  a  Binomial. 

We  find  by  actual  multiplication, 

(a  +  by  =  {a  +  h)  X  {a  +  b)  =0?  +  2  ah  -\-  h\  (1) 

{a-hf={a-h)  x  {a-h)  =  o? -2  ab-\-V\  (2) 

That  is, 

The  square  of  tJie  Bum  of  two  numbers  equals  the  square  of 
the  first,  plus  twice  the  product  of  the  fii^st  by  the  second,  plus  the 
square  of  the  second. 

The  square  of  the  difference  of  two  numbers  equals  the  square 
of  the  first,  minus  twice  the  product  of  the  first  by  the  second, 
plus  the  square  of  the  second. 

In  the  remainder  of  the  book,  we  shall,  for  the  sake  of  brevity,  use  the 
expression  "the  difference  of  a  and  &"  to  denote  the  remainder  obtained 
by  subtracting  b  from  a. 

1.  Square  3  0.2 -2  6. 

By  (2),  (3  a^  -  2  bf  =^  (3  a'f  -  2  (3  a')  (2  b)  +  (2  bf 

=  9  a'- 12  a'b-^- 4.  b'  (§  130). 

If  the  first  term  of  the  binomial  is  negative,  it  should  be  enclosed, 
negative  sign  and  all,  in  parentheses,  before  applying  the  rule. 

2.  Square  -  4  a^  +  9. 

(-  4  ar^  +  9)-  =  [(-  4  x')  +  9]^ 

=.(_4ar^y+2(-4a^)(9)  +  9^by  (1) 
=  16a^-72(^  +  Sl. 

132.  Product  of  the  Sum  and  Difference  of  Two  Numbers. 

We  find  by  actual  multiplication, 

(a-\-b){a-b)  =  a'-b\ 

That  is,  the  product  of  the  sum  and  difference  of  two  numbers 
equals  the  difference  of  their  squares. 

1.    Multiply  Qa  +  ^b""  by  Q>a-5b\ 
By  the  rule, 
(6  a  +  5  W)  (6a-5b')  =  (6  a)'  -  (5  6^)-  =  36  a' -  25  b'\ 


SPECIAL  METHODS  61 

2.  Multiply  -  ic^  +  4  by  -  ic^  -^  4. 

(-x'2  +  4)(-a^-4)  =  [(-.r^)4-4][(-:i^)-4] 
=  (_a;2)2_42=;:a^-,16. 

3.  Expand  (a  +  b  —  c)  (a  —  b  -{-  e). 

To  expand  an  algebraic  expression  is  to  perform  the  operations  indi- 
cated. 

By  §  82,  (a  +  &  -  c)  (a  -  6  +  c)  =  [a  +  (^>  -  c)]  [a  -  (6  -  c)] 

=  ct^  —  (b  —  c)'^,    by  the  rule, 

=  a2  ^  (Z)2  -  2  6c  +  c2) 

=.a2-^62^2  5c-cl 

4.  Expand  (x -{- y  -i- z)  (x  —  y  +  2;). 

(a;  -f-  2/  +  2)  (^  -  2/  +  2^)  =  [(a^  +  2;)  +  2/]  C(^  +  2;)  -  2/] 

=  (x  +  2)^-2/^ 

=  x2  +  2  a^2  +  ^2  _  ^2^ 

133.   Product  of  Two  Binomials  having  the  Same  First  Term. 

We  find  by  actual  multiplication 

(x  -\- a) (x -\^ b)  =  x^ -{-{a  -\-b)x  +  ab. 
That  is. 

The  product  of  tivo  binomials  having  the  sayne  first  term  equals 
the  square  of  the  first  term,  plus  the  algebraic  sum  of  the  second 
terms  multiplied  by  the  first  term,  plus  the  product  of  the  second 
terms. 

1.  Multiply  £c  — 5  by  a; +  3. 

By  the  above  rule,  the  coefficient  of  x  is  the  sum  of  —  5  and 
+  o,  or  —  2,  and  the  last  term  is  the  product  of  —  5  and  +  3, 
or  - 15. 

Whence,  (a;  -  5)  (a;  +  3)  =  a^  -  2  a;  ^  15. 

2.  Multiply  x  —  ^  by  a;  — 3. 

The  coefficient  of  x  is  the  sum  of  —  5  and  —  3,  or  —  8,  and 
the  last  term  is  the  product  of  —  5  and  —  3,  or  15. 

Whence,  (a;  -  5)  (a;  -  3)  =  a;^  -  8  ic  + 15. 


62        ADVANCED  COURSE  IN  ALGEBRA 

3.  Multiply  ab  —  4:  by  ab-{-7. 

By  the  rule,     (ab  -  4)  (ab  +  7)  =  a%^  +  3  a&  -  28. 

4.  Multiply  m  +  n -\- 6  hj  m -i- n -{- S. 

(m  +  71  +  6)  (?>i  +  ii  +  8)  3=  [(m  +  n)  +  6]  [(m  +  7i)  +  8] 
=  (m  +  ny  + 14  (m  +  n)  +  48. 
134r.   Square  of  a  Polynomial. 

By  §  131, ,  (1),  (a,  +  aa)^  =  a,'  +  a/  +  2  a^a^.  (1) 

We  also  have, 

(tti  +  ^2  +  «3)^  =[(«!  + ^2)  +  «3]^ 

=  («!  +a2)2  +  2  («!  +  as)  X  as  +  ag^ 

=  a/  +  2  aitta  +  «2^  +  2  ajag  +  2  agOEg  +  a^^ 

=  ai  +  ai  -\-a^  -[-2  a^a^  +  2  a^a^  +  2  a2<^3-         (2) 

The  results  (1)  and  (2)  are  in  accordance  with  the  following 
law: 

The  square  of  a  polynomial  equals  the  sum  of  the  sqiiares 
of  its  terms,  plus  twice  the  product  of  each  term  by  each  'of  the 
following  terms. 

We  will  now  prove  that  this  law  holds  for  the  square  of  any 
polynomial. 

Assume  that  the  law  holds  for  the  square  of  a  polynomial  of 
m  terms,  where  m  is  any  positive  integer ;  that  is, 

(%  +  tto  +  «3  +  •••  +  a«-i  +  ^mf 
=  a^^  J^  ai  -{-'"  +  aj  +  2  a^  (a2+. ..  +  «„.) 

+  2  a2(a3+  -  +a.)  +  •••  +  2  a,,_ia^.  (3) 

Then,    (a^  +  ag  +  %  H h  ot^  +  «m+i)^ 

=  [(ai  4-  a,  4-  . . .  +  a  J  +  a^^{\\ 
=  (ai  +  «2H h  a^y 

+  2  (ai  +  a,  H h  a  J  a^+i  +  a^+i^,  by  (1) 

=  af  +  a^^  +  •  ••  +  a^'  +  a^+i' 

+  2ai(a2H h  ««  +  a^+i) 

+  2  (I2  («3  +  •••  +  a,.  +  f^m+i)  +  •••  +  2  a^a^+i,  by  (3). 


SPECIAL  METHODS  63 

This  result  is  in  accordance  with  the  above  law. 

Hence,  if  the  law  holds  for  the  square  of  a  polynomial  of  m 
terms,  where  m  is  any  positive  integer,  it  also  holds  for  the 
square  of  a  polynomial  of  m  -f- 1  terms. 

But  we  know  that  the  law  holds  for  the  square  of  a  polyno- 
mial of  three  terms,  and  therefore  it  holds  for  the  square  of  a 
polynomial  of  four  terms ;  and  since  it  holds  fc^i-  the  square  of 
a  polynomial  of  four  terms,  it  also  holds  for  the  square  of  a 
polynomial  of  five  terms ;  and  so  on. 

Hence,  the  law  holds  for  the  square  of  any  polynomial. 

The  above  method  of  proof  is  known  as  Mathematical  Induction. 

Ex.   Expand  (2x'-Sx-  5)1 
In  accordance  with  the  law,  we  have 
(2  aj2  -  3  a;  -  5)2 
=  (2x^'  +  {-^xf+{-5y 

+  2  (2  x")  (-  3  aj)  +  2  (2  a?)  (_  5)  +  2  (-  3  x)  (-  5) 
=  4  a;*  +  9  a^  +  25  -  12  a^  -  20  x-^  +  30  a; 
=  4.x*  -12  ^  -11  ix?  -{-  2,0  X  +  2^. 

135.   Cube  of  a  Binomial. 

We  find  by  actual  multiplication, 

(a  +  6)3  =  a^  -I-  3  a'b  +  3  aft^  +  W,  (1) 

(a  _  &)3  =  a^  _  3  a?h  -f  3  aft^  _  53^  (2) 

That  is. 

The  cube  of  the  sum  of  two  numbers  equals  the  cube  of  the 
first,  plus  three  times  the  square  of  the  first  times  the  seco7id,  plus 
three  times  the  first  times  the  square  of  the  second,  plus  the  cube 
of  the  second. 

The  cube  of  the  difference  of  two  numbers  equals  the  cube 
of  the  first,  minus  three  times  the  square  of  the  first  times  the 
second,  plus  three  times  the  first  times  the  square  of  the  second, 
minus  the  cube  of  the  second. 


64  ADVANCED  COURSE   IN   ALGEBRA 

1.  Find  the  cube  of  a-h^  b. 

By  (1),  (a  +  2  bf  =  a«  +  3  aX2  &)  +  3  a  (2  by  +  (2  bf 
=  a^  +  6a'b  +  12  ab'  +  8  b\ 

2.  Find  the  cube  of  2  a?^  —  d  /. 

By  (2),  (2a;^-5  2/^)« 

=  (2  a^.)«  -  3  (2  a!«)X5  /)  +  3  (2  a^)  (5  /)2  -  (5  y^ 
=  8  a;^  -  60  xY  +  150  xY  - 125  /. 

136.   Cube  of  a  Polynomial. 

By  §  135,  (1),  (ai  -f  ^s)'  =  ^i'  +  ^2'  +  3  a^%  +  3  a^a^l  (1) 

We  also  have,  (ai  +  aa  +  ^3)^ 
=  [(ai  +  a2)  +  a3]3 

=  (%  +  ttg)^  +  3  («!  +  a2)^«3  +  3  (oi  +  a2)ot3^  +  «3^ 
=  ai^  +  3  ai\«2  +  3  aia^^  +  cxg^  +  3  aj^aa  +  6  aia2a3  ■ 

+  3  a2%  +  3  ciiag^  +  3  a2«3^  +  «3^ 
=  tti^  +  ag^  +  aa^  +  3  ai%  4-  3  ai%  +  3  a2^«i  +  3  a2% 

+  3  ttg^iofi  4-  3  a/ag  +  6  ai«2«3-  (2) 

The  results  (1)  and  (2)  are  in  accordance  with  the  following 
law: 

The  cube  of  a  polynomial  equals  the  sum  of  the  cubes  of  its 
terms,  plus  three  times  the  product  of  the  square  of  each  term 
by  each  of  the  other  terms,  plus  six  times  the  product  of  every 
three  different  terms. 

We  will  now  prove  by  Mathematical  Induction  (see  §  134), 
that  this  law  holds  for  the  cube  of  any  polynomial. 

Assume  that  th^  law  bolds  for  the  cube  of  a  polynomial  of  m 
terms,  where  m  is  any  positive  integer ;  that  is, 

(«!  +  «2  +  ag  +  •  • .  +  a^_2  +  a^-i  +  «m)^ 

z=:za^  +  ai-\ VaJ 

4-  3  a^ia.^  +  ag  +  •••  4-  «^)  +  3  a^Xa^  4-  «3  H h  « J 

4-  •••  4-  3  aj(a^  +  ag  4-  •••  +  a^^-x) 

4-  6  ttjaattg  4-  •  •  •  4-6  a^_2«m-i<^w  (3) 


SPECIAL   METHODS  Qd 

Then,  (a^^  a^  +  a^-h  •-  +  a^_i  +  a,,  +  a«+i)^ 
=  [(tti  +  as  +  ag  +  . . .  +  a^_i  +  a  J  +  a^+^f 
=   («i  +  aa  +  a3  +  •..  4-  of,„_i  +  a^f 

+  3  (ai  +  0!2  +  %  H +  «^_i  4-  a  J^«^+i 

4-3(ai  +  a2+«3+  •••  +an,_,  +  a^)a,,J+a^^,' (^  135). 
Then,  by  (3)  and  §  134, 
(«!  +  as  +  «3  +  •••  +  cim-i  +  a^  +  a^+i)8 
=  ai3  +  a2^H \-aJ 

+  3ai2(a2  +  «3+  •••  4-a^) 
+  3a2^(ai  +  a3+  ...  +aj+  ... 
4-  3  a  J(ai  +  ag  +  .  •  •  +  a^_i) 
+  6  a^a^a^  +  ...  +6  a^.ga^.^a^ 

+  3  a,„+i(ai2  +  as^  +  ...  +  aj  +  2  aiag H \-2  a^a^ 

+  2  a^ttg  H +2  a^a^  +  ...  +  2  a„_iaj 

+  3  a^+i^(ai  +  «2  +  «3  +  •••  +  ttm)  +  a^+i 
or,  ai«  +  a2«+  ...  +aj  +  a.+i' 

+  3  ai2(a2  +  •  •  •  H-  «^  +  a^+i) 
4-  3  a22(ai  +  ^3  +  •••  -f  ^m+i)  +  •- 
+  3  a^+i\ai  +  as  4-  ••'  +  «m) 
4-  6  aittsag  4-  . . .  +  6  a^_ia^a,„+i. 

This  result  is  in  accordance  with  the  above  law. 

Hence,  if  the  law  holds  for  the  cube  of  a  polynomial  of  m 
terms,  where  m  is  any  positive  integer,  it  also  holds  for  the 
cube  of  a  polynomial  of  m  + 1  terms. 

But  we  know  that  the  law  holds  for  the  cube  of  a  polynomial 
of  three  terms,  and  therefore  it  holds  for  the  cube  of  a  poly- 
nomial of  four  terms ;  and  since  it  holds  for  the  cube  of,  a 
polynomial  of  four  terms,  it  also  holds  for  the  cube  of  a  poly- 
nomial of  five  terms  ;  and  so  on. 

Hence,  the  law  holds  for  the  cube  of  any  polynomial. 

Ex.     Expand  (2  ar^  -  ar' 4- 2  x  -  3)^ 


66  ADVANCED   COURSE   IN   ALGEBRA 

In  accordance  with  the  above  law,  we  have 
(2  a^3  _  ^  _^  2  «  -  Sy. 

=  {2a^y-]-(-xy-^(2xy-{-(-3f  +  3(2a:^)X~x'  +  2x-3) 

-f-  3  ( -  ^2)2(2  :ij3  +  2  a;  -  3)  +  3  (2  xy(2  a^-x'-  3) 

-i-S(-3y(2a^-x''-\-2x) 

+  6(2  a^)  (-  x")  (2  x)  +  6(2  a^)  (-  x')  (-  3) 

+  6(2  0^)  (2  oj)  (-  3)  +  6(-  x')  (2  a;)  (-  3) 
=  Sx'-x^  +  Sx'-2T-12  x^  +  24  .^^-36  x^+Q,  x' +  (S  x^-^  x' 

+  24  0^  -  12  x^  -  36  i»2  _^  54  ar^  -  27  x^  +  54  a^  -  24  x^ 

+  36  a^  -  72  a;^  +  36  of 
=  S  x^  -12  0^  +  30  x'  -61  x"  +  66  a^  -  93  x*  +  98  a:^  -  63  a^ 

+  54.X-27. 

EXERCISE   6 

Write  by  inspection  the  values  of  the  following  : 
1.    (6a3aj2)3.  2.    (- i  ab^c^)"^.  3.    (-Sx^yz^y. 

4.  (3  +  7x2)2.  7.    (-Ox?/ -11x0)2. 

5.  (2a3_5^2c)2.  8.    (8  x?  -9x^)2;    ^j  and  g  being 

6.  (—  m'^n'^  +  4^9^)2.  positive  integers. 

Write  by  inspection  the  values  of  the  following  : 
9.    (5a2  + i2  63c)(5a2_  I2&3c). 

10.  (  -  10  m%  +  13  x5)  (-  10  m^n  -  13  x^). 

11.  (a2m  4-  x^'>')(a^'»'  —  x^")  ;  w  and  n  being  positive  integers. 

Expand  the  following : 

12.  (a  -  &  +  c)  (a  -  6  -  c).  14.    (x2  -\-xy  +  y^) (x2  -  xy  +  y^). 

13.  (x  +  ?/  +  3)(x-?/-3).  15.    (a2  +  6a-4)(a2_  5a  +  4). 

16.  (4x2  +  3x  +  7)(4x2  +  3x-7). 

17.  (m*  +  5  to2w2  +  2  «4)  (m*  -  5  ?7i2^2  _  2  w^). 

Write  by  inspection  the  values  of  the  following  : 

18.  (x  +  2)(x  +  10).  22.    (wn  + ll)(mn  +  2). 

19.  (x-5)(x  +  7).  23.    (a26  4.3c3)(a2;,_8c3). 

20.  (x-2  -  4)  (x2  -  14),  24.    (a-h  -5)(a-b-9). 

21.  (X4- 7a)(x- 15  a).  25.    (x  +  y  -  6  02)(a:  +  y  +  12  5;2). 


SPECIAL  METHODS  67 

Expand  the  following : 

26.  (3  ic2  +  5  X  -  4)'-^.  28.    (2  x"^  -  3  a:^  +  x  -  2)2. 

27.  {a-h-c  +  ay.  29.    (a*  +  3  a^  _  4  ^^2  _  2  «  +  1)2. 

Write  by  inspection  the  values  of  the  following  : 

30.  (a  +  3  6)3.  32.    (3  oT-h  +  2  c3)3. 

31.  (7x*-x3)3.  33.    (5  wx2  -  4  w2/3)8. 

Expand  the  following : 

34.  (a2  +  a  _  2)3.  36.    (a  -  &  +  c  -  (Z)3. 

35.  (2  x2  -  4  X  +  3)3.  37.    (3  x3  -  4  x2  -  2  x  +  1)3. 

Simplify  the  following : 

38.  (3  a2  +  5  &)2(3  ^2  _  5  5)2,  4I.    (^  +  i)3(^^  _  1)3. 

39.  (x  +  5)(x-2)(x-5)(x  +  2).     42.    (x  +  y  -  zY(x  -  y  +  zy. 

40.  (2  -  x)(2  +  X)  (4  +  x2).  43.    (a  +  6  +  c)3(a  +  6  -  g)3. 

44.  (x  +  y  +  0)2  +  (y  +  0  -  x)2  4-  (0  +  X  -  ?/)2  +  (X  +  ?/  -  z)\ 

45.  (a  +  6  +  c) (6  +  c  -  a) (c  +  «  -  &)  (a  +  &  -  c). 

46.  (m  +  ny  -  (m  -  ny  -  3(m  +  w)2(?7i  -  n)  +  3(m  +  n){m  -  tiy. 

137.  We  find  by  actual  division, 

tj=L^  =  a-h.     (1)  t^l^  =  a  +  h.     (2) 

That  is,  ^' 

i/"  the  difference  of  the  squares  of  two  nurtibers  he  divided  by  the 
su7n  of  the  numbers,  the  quotient  is  the  difference  of  the  numbers. 

If  the  difference  of  the  squares  oftivo  numbers  be  divided  by  the 
difference  of  the  numbers,  the  quotient  is  the  sum  of  the  numbers. 

Ex.   Divide  25  ?/V  -  9  by  5  yz"  -}-  3. 
By  §  130,  25  ?/V  is  the  square  of  5  yzK 

Then,  by  (1),       ^^  •^'^'  ~  ^  =  5yz'-  3. 

138.  We  find  by  actual  division, 

t±J^=::a:'-ab  +  b\     (1)  t:Z-^  =  a^^ab-^b\     (2) 

a+b  a—b 


68  ADVANCED   COURSE   IN   ALGEBRA 

That  is, 

If  the  sum  of  the  cubes  of  two  numbers  be  divided  by  the  sum 
of  the  numbers,  the  quotient  is  the  square  of  the  first  number, 
minus  the  product  of  the  first  by  the  second,  plus  the  square  of  the 
second  number. 

If  the  difference  of  the  cubes  of  tivo  numbers  be  divided  by  the 
difference  of  the  numbers,  the  quotient  is  the  square  of  the  first 
number,  plus  the  2oroduct  of  the  first  by  the  second,  plus  the  square 
of  the  second  number. 

Ex.   Divide  27  a^  -Whj  3a-  b. 
By  §  130,  27  a^  is  the  cube  of  3  a. 

Then,  by  (2),      ^^  ^^' ~  ^'  ==  9  a^  +  3  ab  +  b\ 
3  a—  b 

139.   The  Remainder  Theorem. 

Let  it  be  required  to  divide  2  x^  —  7  x^  -\- 10  x  —  3  hj  x  —  2. 

x-2 


2a^- 

-  7  a.'2  +  10  a;  - 

-3 

2a^- 

-3  of 

-3x'  + 

6^ 

4:X 

» 

4tx- 

-8 

2x^-3x-{-^ 


5 
The  division  is  not  exact,  and  there  is  a  final  remainder  5. 
Now  if  we  substitute  2  for  x  in  the  dividend,  we  have 

2  X  2^  -  7  X  2^  +  10  X  2  -  3,  which  equals  5. 
This  exemplifies  the  following  law : 

If  any  rational  integral  polynomial,  involving  x,  be  not  divisible 
by  X  —  a,  the  remainder  of  the  division  equals  the  result  obtained 
by  substituting  a  for  x  in  the  given  polynomial. 

The  above  is  called  The  Remainder  Theorem. 
To  prove  the  theorem,  let  D  be  any  rational  integral  poly- 
nomial, involving  x,  not  divisible  hj  x  —  a. 


SPECIAL  METHODS  69 

Let  the  division  be  carried  out  until  a  remainder  is  obtained 
which  does  not  contain  x. 

Let  Q  denote  the  quotient,  and  R  the  remainder. 

Since  the  dividend  equals  the  product  of  the  divisor  and 
quotient,  plus  the  remainder,  we  have 

Q{x-a)+R==D. 

Substitute  in  this  equation  a  for  x. 

The  term  Q{x  —  a)  becomes  zero ;  and  since  R  does  not  con- 
tain X,  it  is  not  changed,  whatever  value  is  given  to  x. 

Then,  R  must  equal  the  result  obtained  by  substituting  a 
for  X  in  D. 

140.  The  Factor  Theorem. 

If  any  rational  integral  polynomial,  involving  x,  becomes  zero 
when  X  is  put  equal  to  a,  the  polynomial  has  si;  —  a  as  a  factor. 

For  by  §  139,  the  remainder  obtained  by  dividing  the  poly- 
nomial by  a;  —  a  is  zero. 

141.  It  follows  from  §  140  that. 

If  any  rational  integral  polynomial,  involving  x,  becomes  zero 
when  X  is  put  equal  to  —  a,  the  polynomial  has  x  -i-a  as  a  factor. 

142.  We  will  now  prove  that,  if  n  is  any  positive  integer, 
I.   a"  —  b""  is  always  divisible  by  a—  b: 

II.    d^  —  5"  is  divisible  bya-{-bifn  is  even. 

III.  a"  +  b""  is  divisible  bya-^bifn  is  odd. 

IV.  a"  4-  &"  is  divisible  by  neither  a-^b  nw  a  —  b  ifn  is  even. 
Proof  of  1. 

If  b  be\  substituted  for  a  in  a*"  —  ?>",  the  result  is  6"  —  6",  or  0. 
Then  by  §  140,  a"  —  6"  has  a  —  6  as  a  factor. 

Proof  of  11. 

If  —  6  be  substituted  for  a  in  a**  —  6",  the  result  is  (  —  6)"—  6" ; 
or,  since  n  is  even,  6"  —  6",  or  0. 

Then  by  §  141,  a'^—b""  has  a  +  &  as  a  factor. 


70  ADVANCED   COURSE   IN   ALGEBKA 

Proof  of  111. 

If  —  6  be  substituted  for  a  in  a""  -\-  b"",  the  result  is  (—  6)"+  6"; 
or,  since  n  is  odd,  —  6"  +  b"",  or  0. 
Then,  a^  +  6"  has  a  +  6  as  a  factor. 

Proof  of  lY. 

If  —  6  or  +  6  be  substituted  for  a  in  a"  +  b"",  the  results  are 
{—by-\-  b""  or  b""  +  6",  respectively. 

Since  n  is  even,  neither  of  these  is  zero. 

Then,  neither  a-\-b  nor  a  —  &  is  a  factor  of  a"  +  6". 

143.  We  find  by  actual  division 
=  a^-  a'b  4-  «&'  -  &^ 

=  a^  +  «'^  +  ab'  +  &', 

=  a^  _  a^ft  +  a'b^  -  aW  +  b\ 

=  a^  +  a^6  +  a'^^  +  aW  +  ^' ;  etc. 
a  —  b 

In  these  results,  we  observe  the  following  laws : 

I.  The  exponent  of  a  in  the  first  term  of  the  quotient  is  less  by 
1  than  its  exponent  in  the  dividend,  and  decreases  by  1  in  each 
succeediyig  term. 

II.  The  exponent  of  b  in  the  secoyid  term  of  the  quotient  is  1, 
and  increases  by  1  in  each  succeeding  term. 

III.  If  the  divisor  is  a—b,  all  the  terms  of  the  quotient  are 
positive ;  'if  the  divisor  is  a-\-b,  the  terms  of  the  quotient  are 
alternately  positive  and  negative. 

144.  We  will  now  prove,  by  Mathematical  Induction,  that  the 
laws  of  §  143  hold  universally. 

Assume  the  laws  to  hold  for  ,  where  n  is  any  positive 

iiitesrer.  ^  ~ 


a' 

-b' 

a 

+  b 

a' 

-b' 

a 

-b 

a' 

+  b' 

a 

+  b 

a' 

-¥ 

r 


SPECIAL  METHODS  71 

Then,  ^^Ll^  =  a--i  4.  a'^-'-b  +  a^-'h'  +  •  •  •  +  6""'.  (1) 

a  —  b 

Now,  = ■ 

a  —  b  a  —  b 

a  —  b 
=  a"  +  ^(a""^  +  a^'-^b  +  a^-^^^  _| ^  ^n-i^^  by  (^^^^ 

=  a'*  +  a^'-'^b  +  «""■&"  H h  &"• 

This  result  is  in  accordance  with  the  laws  of  §  143. 

Hence,  if  the  laws  hold  for  the  quotient  of  the  difference  of 
two  like  powers  of  a  and  b  divided  hy  a  —  b,  they  also  hold  for 
the  quotient  of  the  difference  of  the  next  higher  powers  of  a 
and  b  divided  hy  a  —  b.  .      , . 

•^  qO   TyO 

But  we  know  that  they  hold  for  ,  and  therefore  they 

a  —  b 

hold  for  ^LJZ^.  and  since  they  hold  for  9^^^  they  hold  for 
a—b  a—b 

a'  -V         1 

;  and  so  on. 

a  —  b 

f^n  7^n 

Hence,  the  laws  hold  for  ,  where  n  is  any  positive 

.   ,  a  —  b 

integer. 

Putting  —  b  for  b  in  (1),  we  have 

a-_(      6)n  ^  ^^_^  _^  _  ^^  _^  ,,,  _^  ^_  ^y.,^ 

a-(-b) 

If  n  is  even,  (-  by  =  b%  and  (-  by-^  =  -  b^-^  (§  130). 

Whence,  "^^^^^  =  a^'^  -  a^-'b  +  a-'%' b--\  (2) 

a  +  b 

If  n  is  odd,  (-  6)"  =  -  6^  and  (-  6)"-^  =  +  &""'. 

Whence,  ^!±_^  =  a^-^  -  a"-'6  +  a^-^ft^ ^  ^n-i^  (3) 

Equations  (2)  and  (3)  are  in  accordance  with  the  laws  of 
§  143. 


72  ADVANCED   COURSE   IN   ALGEBRA 

op, tm 

Hence,  the  laws  hold  for  ,  where  n  is  any  even  posi- 

a-j-  5 

^w  _j_  ^n 

tive   integer,  and  for   — — — ,  where  n  is   any  odd  positive 

.    .  a  +  6 

integer. 

145.     1.   Divide  a^  -  6^  by  a  -  6. 

Bv  §  143,  tjul-  =  a«  +  a^6  +  ^452  ^  ^3^3  j^  ^254  _^  ^^5  _|_  ^^^ 
a  —  h 

2.   Divide  16  0^4-81  by  2  aj  + 3. 
By  §130,  Ux'={2x)\ 

Then,  l|^Il|l=(2a^)3_(2aj)2.3  +  (2a;).32-33 

=  8  a^  - 12  a?^  _^  18  a;  -  27. 

The  absolute  value  of  any  term  after  the  first,  in  equations  (1),  (2), 
and  (3),  of  §  144,  may  be  obtained  by  dividing  the  absolute  value  of  the 
preceding  term  by  a,  and  multiplying  the  result  by  h. 

This  would  be  the  shortest  method  if  the  numbers  involved  were  large. 

EXERCISE  7 


Write  by  inspection  the  values  of  the  following : 
36  a^  -  49 


5. 


R 

216  mhi^  +  343  p^ 

6  TO«2  +  7  p3 

7. 

a* -64 

a  +  6 

8, 

m^  —  n^ 

m  ~  n 

9. 

1-x 

0. 

a^  +  x^ 

11. 


6a +  7  6  TO«2  +  7  p3  n  -  X 

2    121  x^  -  64  2/4^2  ^  q4  _  54  ^^    gS  -  64  56 

11x3-  8  2/20  '  '  a  +  6  *                           '     a-26    * 

w3  -  1  o  m^  -  n^  10    ^^25  m4  -  256 

n  —  1  m  ~  n  5m  —  4 

8  +  m6  «  1  -  x6  j^    256  a8  _  a;8 


2  +  m2  1-x  2a  +  x 

125  gg  -  27  x8  j^j    «L±^.  15    243  x^  +  1024  y^^ 

5a2_3a;  'oj^a;  *         3x  +  4y 


146.   Symmetry. 

An  expression  containing  two  or  more  letters  is  said  to  be 
symmetrical  with  respect  to  any  two  of  them,  when  they  can 
be  interchanged  without  altering  the  value  of  the  expression. 

Thus,  a  +  &  +  c  is  symmetrical  with  respect  to  a  and  h  ;  for,  on  inter- 
changing these  letters,  the  expression  becomes  &  +  a  +  c. 


SPECIAL  METHODS  73 

An  expression  containing  three  or  more  letters  is  said  to  be 
symmetrical  with  respect  to  them  when  it  is  symmetrical  with 
respect  to  any  two  of  them. 

Thus,  ab  -\-  bc  +  ca  is  symmetrical  with  respect  to  the  letters  a,  6,  and 
c  ;  for  if  a  and  b  be  interchanged,  the  expression  becomes  ba  -{-  ac-}-  cb^ 
which  is  equal  to  a6  +  6c  +  ca. 

And,  in  like  manner,  ab  +  be  ■}■  ca  is  symmetrical  with  respect  to  b  and 
c,  and  with  respect  to  c  and  a. 

147.  Cyclo-symmetry. 

An  expression  containing  n  letters,  a,  h,  c,  •••,  m,  n,  is  said 
to  be  cy do-symmetrical  with  respect  to  them  when,  if  a  is  sub- 
stituted for  &,  6  for  c,  •••,  m  for  n,  and  n  for  a,  the  value  of  the 
expression  is  not  changed. 

The  above  is  called  a  cyclical  interchange  of  letters. 

Thus,  (a  —  b)(b  —  c)(c  —  a)  is  cyclo-symmetrical  with  respect  to  a,  6, 
and  c ;  for  if  a  is  substituted  for  b,  b  for  c,  and  c  for  a,  it  becomes 
(c  —  a)(a  —  6)  (b  —  c),  which  by  the  Commutative  Law  for  Multiplication 
is  equal  to  (a  —  6)(6  —  c)(c  —  a). 

148.  Every  expression  which  is  symmetrical  with  respect  to 
a  set  of  letters  is  also  cyclo-symmetrical  with  respect  to  them. 

For  since  any  two  letters  can  be  interchanged  without  alter- 
ing the  value  of  the  expression,  the  condition  for  cyclo-sym- 
metry will  be  satisfied. 

But  it  is  not  necessarily  true  that  an  expression  which  is 
cyclo-symmetrical  with  respect  to  a  set  of  letters  is  also  sym- 
metrical with  respect  to  them  ;  for  it  does  not  follow  that  any 
two  letters  can  be  interchanged  without  altering  the  value  of 
the  expression. 

149.  It  follows  from  §§  146  and  147  that,  if  two  expressions 
are  symmetrical  or  cyclo-symmetrical,  the  results  obtained  by 
adding,  subtracting,  multiplying,  or  dividing  them  are,  respec- 
tively, symmetrical  or  cyclo-symmetrical. 

150.  Applications. 

The  principle  of  symmetry  is  often  useful  in  abridging  alge- 
braic operations. 


74  ADVANCED   COURSE   m  ALGEBRA 

1.  Expand  (a  +  6  +  c)l 

We  have,  (a  +  ^  +  c)^  =  (a  +  ^  +  c)  (a  +  &  +  c)  (a  +  6  +  c). 

This  expression  is  symmetrical  with  respect  to  a,  b,  and  c 
(§  146),  and  of  the  third  degree. 

There  are  three  possible  types  of  terms  of  the  third  degree 
in  a,  b,  and  c;  terms  like  a^,  terms  like  a^b,  and  terms  like  abc. 

It  is  evident  that  a^  has  the  coefficient  1 ;  and  so,  by  sym- 
metry, b^  and  c^  have  the  coefficient  1. 

It  IS  evident  that  a^b  has  the  coefficient  3 ;  and  so,  by  sym- 
metry, have  b-a,  b~c,  c^b,  c^a,  and  a^c. 

Let  m  denote  the  coefficient  of  abc. 

Then,  (a  +  b  +  cf 

=  a^  4_  ^3  +  c^  +  3  (a26  -f  b"-a  +  bh  +  c~b  +  (?a^  ah)  +  mabc. 

To  determine  m,  we  observe  that  the  above  equation  holds 
for  all  values  of  a,  6,  and  c. 

We  may  therefore  let  a  =  6  =  c  =  1. 

Then,   27  =  3  +  18  +  m ;  and  m  =  Q. 

Whence,  (a  +  6  +  c)^ 
=  a^  +  ?>3  _^  c^  _^  3  (a-'ft  +  &'a  +  bh  +  0^6  +  c^a  -f-  ah)  +  6  a&c. 

The  above  result  may  be  written  in  a  more  compact  form  by  represent- 
ing the  sum  of  terms  of  the  same  type  by  the  symbol  S  ;  read  sigma. 

Thus,  (2a)5  =  Sa3  +  3  ^a%  +  6  abc. 

2.  Expand  {x  —y —  z)^ -\-{ij —  z —  x)""  +  {z  —  x  —  yf. 

This  expression  is  symmetrical  with  respect  to  x,  y,  and  z, 
and  of  the  second  degree. 

The  possible  types  of  terms  of  the  second  degree  in  x,  y,  and 
z  are  terms  like  x^,  and  terms  like  xy. 

It  is  evident,  by  the  law  of  §  134,  that  x^  has  the  coefficient 
3 ;  and  so,  by  symmetry,  have  y"^  and  z-. 

Let  m  denote  the  coefficient  of  xy. 

Then,  {x-y -zf -\- {y -z-xf -^  {z- x- yf 
=  3  (a^  -f  2/^  +  2^)  +  m{xy  -{-  yz  -f-  zx). 

To  determine  m,  put  x  =  y  =  z  =  l. 

Then,  3  =  9  -f  3  m,  or  m  =  -  2. 


SPECIAL  METHODS  75 

Whence,  (x  —  y  —  zY-[-{y  —  z  —  xf-\-{z—x  —  yy 

=  3(x'-^y'-^z')-2(xy  +  yz  +  zx). 
3.   Expand 

(a  H-  6  +  c)3  +  (a  +  6  -  c)3  +  (^  +  c  -  a)3  +  (c  +  a  -  b).^ 

The  expression  is  symmetrical  with  respect  to  a,  b,  and  c, 
and  of  the  third  degree. 

The  possible  types  of  terms  are  terms  like  a^,  terms  like  a-6, 
and  terms  like  abc. 

It  is  evident,  by  the  law  of  §  136,  that  a?  has  the  coefficient 
2 ;  and  so,  by  symmetry,  have  W  and  cl 

Also,  by  §  136,  o?b  has  the  coefficient  3  +  3  +  3-3,  or  6; 
and  so,  by  symmetry,  have  Wa,  bh,  c^b,  c\  and  ah. 

Again,  abc  has  the  coefficient  %  —  Q  —  Q>  —  Q,  or  —  12. 

Whence, 

(a  +  6  +  c)3  +  (a  +  6  -  c)3  +  (6  +  c  -  a)3  +  (c  +  a  -  bf 
=  2  (a'  +  63  .^  c^)  ^  6  ((^25  _^  ^2^^  _^  ^2^  _l_  ^2^  ^  ^2^  ^  ^2^^)  _  12  a6c. 

EXERCISE    8 

1.  In  the  expansion  of  an  expression  which  is  symmetrical  with  re- 
spect to  a,  &,  and  c,  what  are  the  possible  types  of  terms  of  the  fourth 
degree?  of  the  fifth  degree?   of  the  sixth  degree? 

2.  If  one  term  of  an  expression  which  is  symmetrical  with  respect  to 
a,  b,  and  c,  is  (2 a  —  &  —  c)  (2b  —  c  —  a),  what  are  the  others ? 

3.  Is  the  expression  a(b  —  c)^  +  6  (c  —  a)^  +  c(a  —  b)'^  symmetrical 
with  respect  to  a,  &,  and  c  ? 

4.  Is  the  expression  (x^  —  y^y  +  (y^  —  z^y  -{-  (^2  _  x^y  symmetrical 
with  respect  to  x,  y,  and  z  ? 

Expand  the  following  by  the  symmetrical  method  : 

6.  (a  +  6  +  c)2.  6.    (a  +  &  +  c  +  dy. 

7.  (x-hy-zy+(y  +  z-xy+(z-\-x-  yy. 

8.  (2  a  -  3  &  -  4  c)-^  +  (2  &  -  3  c  -  4  a)2  +  (2  c  -  3  a  -  4  &)2. 

9.  (a  +  6  +  c  -  d)2  +  (6  +  c  +  cZ  -  a)2  +  (c  +  cZ  +  a  -  6)2 

+  (d  +  a  +  6  -  c)2. 

10.  (a  +  6  +  c  +  d)3. 

11.  (a  +  6  +  c)3  +  (a  -  6  -  c)3  +  (&  -  c  -  a)3  +  (c  -  a  -  by. 

12.  (X  +  ?/  -  ^)   (y  +  ;3  -  X)   (5!  +  cc  -  I/). 

13.  [x-^ -^  y^- +  z'2  ^  2  (xy  +  yz -\- zx)y. 

14.  («  +  ?;  +  c)  (a  +  6  -  c)  (?)  +  €•-«)  («  +  c  -  6). 


76        ADVANCED  COURSE  IN  ALGEBRA 


VIII.    FACTORING 

151.  To  Factor  an  algebraic  expression  is  to  find  two  or 
more  expressions  which,  when  multiplied  together,  shall  pro- 
duce the  given  expression. 

152.  In  the  present  chapter  we  consider  only  the  separation 
of  rational  and  integral  expressions  (§  63),  with  integral  nu- 
merical coefficients,  into  factors  of  the  same  form. 

153.  A  Common  Factor  of  two  or  more  expressions  is  an 
expression  which  will  exactly  divide  each  of  them. 

FACTORING 

154.  It  is  not  always  possible  to  factor  an  expression ;  there 
are,  however,  certain  forms  which  can  always  be  factored; 
these  will  be  considered  in  the  present  treatise. 

155.  Case  I.  When  the  terms  of  the  expression  ho>ve  a  com- 
mon  factor. 

1.  Factor  14  0,6*- 35  a^^l 

Each  term  contains  the  monomial  factor  7  aV^. 
Dividing  the  expression  by  7  aW,  we  have  2W  —  6a\ 
Then,  14  ab^  -  35  a^h''  =  7  ab'  {2  b' -5  a'). 

2.  Factor  (2  m  +  3)  a;^  +  (2  m  +  3)  if. 

The  terms  have  the  common  binomial  factor  2  m  -f  3. 
Dividing  the  expression  by  2  m  +  3,  we  have  x'  -f-  y^. 
Then,  (2m-f  3)a;2  +  (2m  +  3)2/'=(2m+3)  (x'-\-f). 

3.  Factor  (a  —  b)m+{b—a)  n. 

By  §  81,  5  _  a  =  -  (a  -  b). 

Then,    (a  —  b)m-]-(b  —  a)n=(a  —  b)m  —  (a  —  b)n 

=  (a  —  b)  (m  —  n). 

4.  Factor  5  a  (x—y)  —3a(x-\-y). 


FACTORING  77 

5a(x-y)-3a(x-{-y)  =  al5(x-y)-3(x-\-y)'] 
=  a  (5  X  —  5  y  —  3  X  —3  y) 
=  a{2x-8y) 
=  2  a  (a;  —  4  y). 

We  may  also  solve  Ex.  3  by  writing  the  first  term  in  the  form 

—  (6  —  a)  m. 
Thus,  (a  —  6)  TO  +  (6  —  a)  w  =  (6  —  a)  n  —  (h  —  a)m 

=  (h  —  a)(n  —  m). 
This  agrees  with  §  91 ;   for,  by  §  91,  the  signs  of  tv^o  factors  of  a 
product  may  be  changed  without  altering  the  value  of  the  expression. 
We  may  thus  have  more  than  one  form  for  the  factors  of  an  expression. 

156.  The  terms  of  a  polynomial  may  sometimes  be  so  ar- 
ranged as  to  show  a  common  polynomial  factor;  and  the 
expression  can  then  be  factored  as  in  §  155. 

1.  Factor  ah  —  ay-\-bx  —  xy. 

By  §  155,  ab  —  ay-j-bx  —  xy  =  a(b  —  y)-\-  x(b  —  y). 
The  terms  now  have  the  common  factor  b  —  y. 
Whence,    ab  —  ay-\-bx  —  xy=  (a  -i-x)  (b  —  y). 

2.  Factor  a3  +  2a2-3a-6. 

The  third  term  being  negative,  it  is  convenient  to  enclose 
the  last  two  terms  in  parentheses  preceded  by  a  —  sign. 

Thus,         a?-\-2a?-3a~Q={a?+2a?)-{3a  +  e>) 

=  a2(a  +  2)-3(a  +  2) 
=  (a2-3)(a  +  2). 

EXERCISE  9 

Factor  the  following : 

1.  (3x  +  5)m  +  (3a;  +  5).  3.    x\by -2z)  -  x\2y -\- z). 

2.  (rii  —  n)x -{■  {n  —  m) {y -{■  z) .       4.    4tx{a—h—c)  —  by(h  +  c  —  d), 

5.  («  -  h)  {mP-  +  xz)  -  (a  -  b)  (m^  -  yz). 

6.  (to  -  w)*  -  2  to  (to  -  n)3  +  to2(w  -  n)2. 

7.  Sxy +  12  ay +  10  bx  + lb  ab.        9.    6  -  10  a  +  27  a2  _  45  «». 

8.  TO*  +  6  to3  -  7  TO  -  42.  10.    26ab  -2S  ad -5  be +  7  cd. 

11.  ax  —  ay  +  az  —  bx  +  by  —  hz. 

12.  3  am  —  0  au  +  4  hm  —  ^hn  +  cm  —  2  en. 

13.  ax  -I  ay  —  az  —  bx  —  by  +  bz  +  ex  +  ey  —  ez. 


78  ADVANCED  COURSE   IN  ALGEBRA 

157.  If  an  expression  when  raised  to  the  nth.  power  (n  being 
a  positive  integer)  is  equal  to  another  expression,  the  first 
expression  is  said  to  be  an  nth  Root  of  the  second. 

Thus,  if  a^  =b,  a  is  an  nth  root  of  b. 

158.  The  Radical  Sign,  V,  when  written  before  an  expres- 
sion, indicates  some  root  of  the  expression. 

Thus,     Va''   indicates  a  second,  or  square  root  of  a^ ; 
V  «^  indicates  a  third,  or  cube  root  of  a^ ; 
-\/«^"  indicates  an  nth  root  of  a^" ;  etc. 
The  index  of  a  root  is  the  number  written  over  the  radical 
sign  to  indicate  what  root  of  the  expression  is  taken. 
If  no  index  is  expressed,  the  index  2  is  understood. 
An  even  root  is  one  whose  index  is  an  even  number ;  an  odd 
root  is  one  whose  index  is  an  odd  number. 

159.  A  rational  and  integral  expression  is  said  to  be  a  perfect 
square,  a  perfect  cube,  or,  in  general,  a  perfect  nth  power,  when 
it  has,  respectively,  a  rational  and  integral  square,  cube,  or  nth 
root. 

160.  Since  {2  a'bf  =  ^  a^b^  (§  130),  a  cube  root  of  ^  a%^ 
is  2  a-b. 

Again,  since  (m^y  =  m^,  a  fourth  root  of  m^  is  m?. 
It  is  evident  from  this  that  every  positive  term,  which  is  a 
perfect  ?ith  power,  has  a  positive  nth  root. 
We  shall  call  this  its  principal  nth  root. 

•     We  also  have  (—  m^)*  =  m^ ;  so  that  another  fourth  root  of  m^  is  —  7n^. 

It  is  evident  from  the  above  that  every  positive  term  which  is  a  perfect 
nth  power,  has,  if  n  is  even,  in  addition  to  its  positive  wth  root,  a  negative 
71th.  root  of  the  same  absolute  value. 

In  the  present  chapter,  only  the  principal  nth  root  will  be  considered. 

161.  Since  (-  3  a^)^  =  -  27  «^  (§  130),  a  cube  root  of  -  27  x^ 
is  -  3  a^. 

It  is  evident  from  this  that  every  negative  term,  which  is  a 
perfect  7ith  power,  has  a  yiegative  nth  root. 
We  shall  call  this  its  princijml  nth  root. 


FACTORING  79 

162.  It  will  be  shown  (§  756)  that  a  number  has  two  differ- 
ent square  roots,  three  different  cube  roots,  and,  in  general, 
n  different  nth  roots. 

It  will  be  understood  throughout  the  remainder  of  the  work, 
unless  the  contrary  is  specified,  that  when  we  speak  of  the  nth 
root  of  a  term,  we  mean  the  principal  nth  root. 

163.  Let  n  be  a  positive  integer,  and  a  and  h  two  equal 
perfect  nth.  powers ;  then,  by  Ax.  5,  §  66, 

That  is,  if  two  perfect  nth  powers  are  equal,  their  principal 
nth  roots  are  equal. 

164.  Any  Root  of  a  Power. 

Eequired  the  value  of  a/o^,  where  m  and  n  are  any  positive 
integers. 

By  §  128,  {a'^Y  =  a"•^ 

Then,  by  §  157,  v^a^  =  a*". 

165.  Any  Root  of  a  Product. 

Let  n  be  a  positive  integer,  and  a,  6,  c,  •••,  numbers  which 
are  perfect  nth  powers. 

By  §  157,         {VabG^'y  =  ahC'". 
Also, 

(Va  X  V^  X  Vc  X  ...)"  =  (Va)"  X  (V6)«  X  {</cy  X  •.. 

=  a6c...,  by  §  157. 

Then,  by  §  163,  Vabc^-  =  Va  X  V^  X  ^c  X  —  ; 

for  each  of  these  expressions  is  the  nth  root  of  abc  •••. 

This  simply  means  that  the  principal  nth  root  of  a  product  is  equal  to 
the  product  of  the  principal  nth  roots  of  the  factors. 

166.  Any  Root  of  a  Monomial. 

From  §§  160,  161,  164,  and  165,  we  have  the  following 
rule  for  finding  the  principal  root  of  a  rational  and  integral 
monomial,,  which  is  a  perfect  power  of  the  same  degree  as  the 
index  of  the  required  root :  \ 


80       ADVANCED  COURSE  IN  ALGEBRA 

Extract  the  required  root  of  the  absolute  value  of  the  numerical 
coefficient,  and  divide  the  exponent  of  each  letter  by  the  iyidex  of 
the  required,  root. 

Give  to  every  even  root  of  a  positive  term  the  positive  sign,  and 
to  every  odd  root  of  any  term  the  sign  of  the  term  itself 

1.  Required  the  cube  root  of  64  x^. 
By  the  rule,  V64^  =  4  a:^. 

2.  Required  the  fourth  root  of  81  m^n^l 

■v^SlmV  =  3  mn\ 

3.  Required  the  fifth  root  of  -  32  an'c"^. 


■V -32  aVc'' =  -2  a'b(^. 

167.  It  follows  from  §  131  that  a  rational  and  integral  tri- 
nomial is  a  perfect  square  when  its  first  and  third  terms  are 
perfect  squares,  and  positive,  and  its  second  term  plus  or  minus 
twice  the  product  of  their  square  roots. 

Thus,  4  a:^  4- 12  xy^  +  9  ?/'*  is  a  perfect  square. 

168.  To  find  the  square  root  of  a  trinomial  perfect  square, 
we  reverse  the  rules  of  §  131 : 

Extract  the  square  roots  (§  166)  of  the  first  and  third  terms, 
and  connect  the  results  by  the  sign  of  the  second  term. 

1.   Find  the  square  root  of  4  a^  + 12  ic^/^  +  9  y^. 


By  the  rule,    V4a?+l2^pT9^  =  2 a;  +  3 /. 

The  expression  may  be  written  in  the  form 

(_2a:)2  +  2(-2a;)(-3  2/2)  +  (-3y2)2. 

which  shows  that  (  —  2,x)-\-(—Sy'^),  or  —2x  —  3y^,  is  also  a  square  root. 
But  the  first  form  is  simpler,  and  will  be  used  in  the  examples  of  the 
present  chapter. 

2.   Find  the  square  root  of  m^  —  2mn-\-  r?. 


By  the  rule,         Vm^  —  2mn-^n^=-m  —  n. 

We  may  write  the  expression  in  the  form  n^  —  2  mn  +  w^ ;  in  which 
case,  by  the  rule,  the  square  root  is  n  -  m. 


FACTORING  81 

169.  Case  II.  Wlien  the  expression  is  a  trinomial  perfect 
square. 

1.  Factor  25  a^  +  40  a6^  + 16  6^ 

By  §  168,  the  square  root  of  the  expression  is  5  a  +  4  6^. 
Then,  25  a'  +  40  ab^  + 16  6«  =  (5  a  +  4  b^f. 

2.  Factor  m'*  —  4  m V  +  4  ii''. 

By  §  168,  the  square  root  of  the  expression  is  either  m^  —  2n^, 
or  2n^  —  m^. 

Then,  m^  -  4  m^n^  +  4  n^  =  (m^  -  2  n%  or  (2  n^  -  mf.  (Com- 
pare §  91.) 

3.  Factor  a;2_2aj(2/_^)  +  (2/_0)2. 
We  have       y?  —  2  x{y  —  z)  -\-  {y  —  zf 

=  \x-{y-z)J^{x-y^zf', 
or,  =:[(2/_2;)-a;J  =  (2/-2;-a;)2. 

4.  Factor  _9a*-6a2-l. 

-9a4-6a2-l  =  -(9a^  +  6a2  +  l)=-(3a2  +  l)2. 

EXERCISE  10 

Find  by  inspection  the  values  of  the  following : 

1.    v^- 125^9^3.  2.    \/l6W^.  3.    v/2i3'a5p5^. 

Factor  the  following : 

4.  81  to2  +  144  w  +  64.  7.  -  121  a'^m'^  +  220  a2&2win  -  100  6%2. 

5.  49  w6  -  168  n^x*  +  144  cc8.  8.  9a;2  _  6x(2/+ ^)  +  (?/ +  0)2. 

6.  -  25  a2  -  60  ax  -  36  a;2.  9.  25(a  -  &)2  +  40(a  -  6)c  +  16  c2. 

10.  (a  +  &)2  +  4(a  +  &)(a-&)+4(a-6)2. 

11.  9(a;  +  ?/)2  -  12(x  H-  y)  (x  -  y)  +  4(x  -  y)\ 

170.  Case  III.     Wlien  the  expression  is  in  the  form 

a2  +  62  +  c2  +  2a6+2ac  +  26c. 
By  §  134,  this  expression  is  the  square  of  a-\-b-\-c. 
Ex.     Factor  ^3? +  y^ -{-^^z^—Qxy —  12xz-{-4.yz, 
We  can  write  the  expression  in  the  form 

9  ar^  —  6  a;?/  +  2/^  — 12  0^2;  +  4  ?/2  +  4  2^. 


82  ADVANCED   COURSE   IN   ALGEBRA 

Or,  by  §§  155  and  169, 

(3  a;  _  2/)2  _  4  ^(3  a^- 2/)  +  4  zl 

This  expression  is  the  square  of  {^x  —y)~2z. 
ThviSj9x^-{-y^-^4:Z^-6xy-12xz-^4:yz={tix-y-2zy. 
By  §  91,  we  may  put  the  result  in  the  form  (-3x+2/  +  2  z)^. 

,     ,  „      .  EXERCISE  II 

Factor  the  foUowmg : 

1.  a2  +  62  +  c2  +  2  a6  -  2  ac  -  2  be. 

2.  1  +  25  m2  +  36  w2  -  10  w  +  12  n  -  60  mn. 

3.  a2  +  81  &2  +  16  _  18  a&  -  8  a  +  72  &. 

4.  9  a;2  +  y2  _|.  25  z^ -^  (i  xy  +  30  xz -\-  10  yz. 

6.    36  m2  +  64  w2  +  x2  +  96  mn  -  12  mx  -  16  nx. 

6.  16  a*  +  9  //  +  81  c*  -  24  a2&2  _  72  a2c2  +  54  52^2. 

7.  25  af^  +  49  yi')  +  36  08  _  70  a;^?/^  +  00  x^z*  -  84  y^z*. 

171.  Case  IV.  TF7ien  the  expression  is  the  difference  of  two 
perfect  squares. 

By  §  132,  a2  -  6^  =  (a  +  h){a  -  b). 

Hence,  to  obtain  the  factors,  we  reverse  the  rule  of  §  132 : 

Extract  the  square  root  of  the  first  square,  and  of  the  second 
square;  add  the  results  for  one  factor,  and  subtract  the  second 
result  from  the  first  for  the  other. 

1.  Factor  36a2M-49c«. 

The  square  root  of  30  a'^b^  is  6  ab\  and  of  49  c«  is  7  c«. 
Then,       36  a^b^  -  49  c«  =  (6  aV  +  7  c'^)(6  ab^  -  7  c"). 

2.  Factor  {2x-^yf  -  {x-y)\ 
By  the  rule,     (2  a;  -  3  yf  -{x-  yf 

=:l(2x-Zy)-\-(x-y)-][{2x-^y)-{x-y)-\ 
=  {2x-3y  +  x-y)(2x-3y-x-\-7j) 
=  (Sx~4.yXx-2y). 

A  polynomial  of  more  than  two  terms  may  sometimes  be 
expressed  as  the  difference  of  two  perfect  squares,  and  factored 
by  the  rule  of  Case  IV. 


FACTORING  83 

3.  Factor  2mn-{-m^-l  +  h\ 

The  first,  second,  and  last  terms  may  be  grouped  together  in 
the  order  m^  +  2  mn  +  n^  j  which  expression,  by  §  168,  is  the 
square  of  m  +  n. 

Thus,      2  mn  -f-  m^  —  1  +  n^  =  (m^  -\-  2  mn  -\-n^)  —  1 

=:(m  +  n)2-l 

=  (m  4-  n  -f-  l)(m  +  w  —  1). 

4.  Factor  12y-{-x'-9y- -i. 

■     12y-\-a^-9y^-4:  =  x^-9y^  +  12y-4: 
=  aj2  -  (9  /  -  12  2/  +.4) 
=  x^_(32/-2)^by§168, 
=  [a:  +  (3.v-2)][x-(32/-2)] 
=  (x  +  3y-2){x-Sy  +  2), 

EXERCISE  12 
Factor  the  following  ,- 

1.  196  m^xia  -  289  n^i/w 

2.  36  a2  _  (2  a  -  3)2. 

3.  16(2m-7a:)2-25(3w  +  42/)2. 

4.  4(8a  + 3&)2-9(4a-56)2. 

5.  a2  +  6-2  _  c2  _|.  2  a5. 

6.  x2-  2/2  _  2?/0-2;2. 

7.  6np  +  16?7i2-  9i>2_n2. 

8.  m2  -  2  9nn  +  w2  -  a;2  +  2  xy  -  y^.  » 

9.  16a2- 8a6  + 62- c2_  I0cd-25d2. 

10.    28  x?/ -  36  ^2  ^.  49  2^2  _|.  60  0-25 +  4x2.      ., 

172.   Case  V.      When  the  expression  is  in  the  form 

X*  +  axY  +  y*- 

Certain  trinomials  of  the  above  form  may  be  factored  by 
expressing  them  as  the  difference  of  two  perfect  squares,  and 
then  employing  §  171. 

1.   Factor  a*  +  a'-^^^  _^  5^ 


84  ADVANCED   COURSE   IN   ALGEBRA 

By  §  167,  a  trinomial  is  a  perfect  square  if  its  first  and  last 
terms  are  perfect  squares  and  positive,  and  its  second  term  plus 
or  minus  twice  the  product  of  their  square  roots. 

The  given  expression  can  be  made  a  perfect  square  by  adding 
a^h^  to  its  second  term ;  and  this  can  be  done  if  we  subtract 
a^h'^  from  the  result. 

Thus,  a'  +  a^h''  +  &*  =  (a^  +  2  o?h\+  h')  -  o?h^ 

=  {a"  4-  &'  +  ah){a''  +  &'  -  ah),  by  §  171, 
=  {o?  +  a6  +  h^ia?  -ab-\-  b'). 
2.   Factor  9  x^  -  37  a;2  _|.  4 

The  expression  will  be  a  perfect  square  if  its  second  term  is 
- 12  x'. 
Thus,  9  o;^  -  37  x2  +  4  =  (9  a:^  -  12  aj2  +  4)  -  25  a^  . 
=  (3x'-  2y  -  (5  xf 
=  {Sx'  +  5x-2){3x'-5x-2). 

The  expression  may  also  be  factored  as  follows : 

9  ic*  -  37  xM-  4  =  (9  x^  +  12  a:2  +  4)  -  49  x^ 
=  (3  a;2  +  2)2  -  (7  x)2 
=  (3  a;2  +  7  a;  +  2)  (3  x2  -  7  a;  +  2). 

Several  expressions  in  the  following  set  may  be  factored  in  two  different 
ways. 

The  factoring  of  trinomials  of  the  form  x*  -f  ax^y^  +  ?/^,  when  the 
factors.involve  surds,  will  be  considered  in  §  459. 

EXERCISE  13 

Factor  the  following : 

1.  x*  +  5ic2  +  9.  5.  9  cc*  +  6  a;2^2  +  49  y4. 

2.  a*  -  21  a262  +  36  64.  6.  16  a^  -  81  a^  +  16. 

3.  4-33a:2  +  4x4.  7.  04  +  64^2  +  25  771*. 

4.  25  m*  +  6  »n2n2  +  n*.  8.  49  a*  -  127  a'^x^  +  81  xc*. 

Factor  each  of  the  following  in  two  different  ways  : 

9.   a;4  -  17  x2  +  16.  11.    16  m*  -  104  m'^x'^  +  25  x^. 

10.   9  -  148  a2  +  64  a*.  12.   S6  a^  -  97  a'^m'^  +  S6  m\ 


FACTORING  85 

173.   Case  VI.     When  the  expression  is  in  the  form 

oc^  -{- ax -\- b. 
By  §  133,  £c-  4-  (m  +  n)x  -\-  mn  =  (x  -\-  m)(x  -\-  7i). 

If,  then,  a  trinomial  is  in  the  form  x^  -\-  ax  -{-  b,  and  a  and  b 
are,  respectively,  the  sum  and  product  of  two  numbers,  the 
factors  are  x  plus  one  number  and  x  plus  the  other. 

The  numbers  may  be  found  by  inspection. 

1.  Factor  x^  +  14:X  + 4:5. 

We  find  two  numbers  whose  sum  is  14  and  product  45. 

By  inspection,  we  determine  that  these  numbers  are  9  and  5. 

Whence,        a^  + 14  a;  +  45  =  (a;  +  9)(a;4- 5), 

2.  Factor  x"^  -5x-\-4:. 

We  find  two  numbers  whose  sum  is  —  5  and  product  4. 

Since  the  sum  is  negative,  and  the  product  positive,  the 
numbers  must  both  be  negative. 

By  inspection,  we  determine  that  the  numbers  are  —4 
and  —  1. 

Whence,  a;^  —  5ic  +  4  =  (a;  —  4)(a;  —  1). 

3.  Factor  a;2  +  6a;-16. 

We  find  two  numbers  whose  sum  is  6  and  product  — 16. 

Since  the  sum  is  positive,  and  the  product  negative,  the 
numbers  must  be  of  opposite  sign ;  and  the  positive  number 
must  have  the  greater  absolute  value. 

By  inspection,  we  determine  that  the  numbers  are  +  8 
and  —2. 

Whence,  a^ ^Qx -16  =  {x-\-  S)(x  - 2). 

4.  Factor  x'  -  abx"  -  42  a-b\ 

We  find  two  numbers  whose  sum  is  —  1  and  product  —  42. 

The  numbers  must  be  of  opposite  sign,  and  the  negative 
number  must  have  the  greater  absolute  value. 

By  inspection,  we  determine  that  the  numbers  are  —7 
and  +6. 

Whence,     x^  -  abx"  -  42  a-b-  =  (0^-7  ab){x'  -\-  6  ab). 


86        ADVANCED  COURSE  IN  ALGEBRA 

5.  Factor  l  +  2a-99al 

We  find  two  numbers  whose  sum  is  +  2  and  product  —  99. 
By  inspection,  we  determine  that  the  numbers   are   + 11 
and   —  9. 

Whence,         1  +  2a -99 ^2=  (1  +  11  a)(l -9a). 

If  the  x^  term  is  negative,  the  entire  expression  should  be 
enclosed  in  parentheses  preceded  by  a  —  sign. 

6.  Factor  24  +  5  a;  — ar^. 

We  have,  24  +  5  a;  -  a^  =  -  (a;^  _  5  a;  -  24) 

=  -(x-8)(a7H-3) 
=  (8-aj)(3  +  a;). 

In  case  the  numbers  are  large,  we  may  proceed  as  follows  : 

Required  the  numbers  whose  sum  is  —  26  and  product  —  192. 

One  of  the  numbers  must  be  + ,  and  the  otlier  — . 

Taking  in  order,  beginning  with  the  factors  +1  x  —  192,  all  possible 
pairs  of  factors  of  —192,  of  which  one  is  +  and  the  other  — ,  we  have  : 

+  1  X  -  192. 
+  2  X  -  96. 
+  3  X  -  64. 
+  4x-  48. 
+  6  X  -  32. 
Since  the  sum  of  +  6  and  —  32  is  —  26,  they  are  the  numbers  required. 

EXERCISE  14    - 

Factor  the  following : 

1.  x2  +  18a;  +  56. 

2.  aj2  +  16  a;  -  67. 

3.  a2-10a-76. 

4.  ?/4  -  21 2/2  +  104. 

5.  77-4x-x2. 

6.  84  +  5  n  -  w2. 

7.  1  +  17  ?n  +  70  m2. 

8.  1  +  5  a&  -  14  ^252. 

9.  (x-?/)2-  Voi^x-y)- 
10.    (m- w)2+21(w-?2) 


11. 

(a  +  x)2-28(a  +  a;)  +  192. 

12. 

95_14x4-a;8. 

13. 

105  +  8  w8  -  m^. 

14. 

l  +  36a^2/2  +  68a;2y4. 

15. 

a;6  -  17  x^yz'^  +  72  y'^z^. 

16. 

a2  -  6  a6  -  91  62. 

17. 

a2  +  32amn  +  112w2w2. 

18. 

CCV   +    7  yfiy^z   _    170  Z'^. 

16. 

19. 

a:2-(2w  +  3w)x  +  6win. 

180. 

20. 

a;2— (a  —  h)x  —  ah. 

FACTORING  87 

174.   Ca  SE  VII.      When  the  expression  is  in  the  form 

ax^  -\-hx-\-c. 

If  a  is  a  perfect  square,  and  h  is  divisible  by  Va,  we  may 
factor  the  expression  directly  by  the  method  of  §  173. 

1.  Factor  9  ic2- 18  a; +  5. 

We  have,       9aj2-18a;  +  5  =  (3  a;)2-6(3a;)  +  5. 

We  find  two  numbers  whose  sum  is  —  6,  and  product  5. 

The  numbers  are  —  5  and  —  1. 

Then,  ^  x^-l^x-\-b={Zx-b){^x-l). 

If  h  is  not  divisible  by  Va,  or  if  a  is  not  a  perfect  square,  we 
multiply  and  divide  the  expression  by  a,  which,  by  §  96,  does 
not  change  its  value. 

2.  Factor6aj2  +  5a;-4. 

Multiplying  and  dividing  the  expression  by  6,  we  have 

6^  I  5^     ^_36r>^  +  30x-.24_(6a;y  +  5(6x)-24 
6  6 

The  numbers  are  8  and  —  3. 

Then,  6^  +  5x-4  =  ii^±|M^:^. 

^  X  o 

Dividing  the  first  factor  by  2,  and  the  second  by  3,  we  have 

6a^  +  5aj-4=(3a;  +  4)(2a;-l). 

In  certain  cases,  th"e  coefficient  of  ar^  may  be  made  a  perfect 
square  by  multiplying  by  a  number  less  than  itself. 

3.  ractor8a^  +  26a;2/  +  15  2/^ 
Multiplying  and  dividing  by  2,  we  have 

8  ic^  4-  26  i»?/  + 15  y^  = o 

^(4a;)^  +  13y(4a^)  +  30/ 

2 

^(4a;  +  10y)(4a;  +  3y) 

2 

=  (2ic  +  52/)(4a;  +  32/). 


88  ADVANCED  COURSE  IN   ALGEBRA 

4.   Factor  2  + 5 a;- 3 aj2. 

^(3xY-5(3x)-Q 

-3 
^(3x-6){3x-\-l) 

-3 
=  (2-x)(l-\-3x). 

EXERCISE  15 

Factor  the  following : 

1.  4a;2  +  28a:  +  45.  8.  72  +  7  a:  -  49  x^. 

2.  6  x2  +  X  -  2.           ^  9.  6  -  ic  -  15  x2. 

3.  25  x'^  -  25  mx  -  6  m^.  10.  5  +  9  n^  -  18  w*. 

4.  10  x2  -  39  X  +  14.              .  11.  21  x2  +  23  x?/  +  6  ?/2. 

5.  12x2  +  llx  +  2.  12.  18  x2  -  27  a6x  -  35  «262. 

6.  20  ^2x2  _  23  ax  +  6.  13.  7(a  -  6)2  -  30(a  -  &)+ 8. 

7.  36x2  +  12x-35.  14.  12(x  +  ?/)2  +  17(x  + 2/)- 7. 

15.  14(m- w)2  +  39a(w- w)+10a2. 

16.  acx'^  —(ad  +  &c)x  +  6(Z. 

175.  It  is  not  possible  to  factor  every  expression  of  the  form 
a^  +  ax  +  bhj  the  method  of  §  173. 

Thus,  let  it  be  required  to  factor  x^-\-lSx-{-  35. 

We  have  to  find  two  numbers  whose  sum  is  ^8,  and  product 
35. 

The  only  pairs  of  positive  integral  factors  of  35  are  7  and  5, 
and  35  and  1 ;  and  in  neither  case  is  the  sum  18. 

In  Chap.  XIX  will  be  given  a  general  method  for  factoring 
any  expression  of  the  form  am?  +  6ic  +  c. 

176.  Case  VIII.  When  the  expression  is  the  cube  of  a 
hinomial. 

Ex.     Factor  8  a^  -  36  a^h^  +  54  a6*  -  27  h\ 

We  must  show  that  the  expression  is  in  the  form  of  the  cube 
of  a  binomial,  as  obtained  by  the  rule  of  §  135,  and  find  its 
cube  root. 


P^ACTORING  89 

We  can  write  the  expression  as  follows : 

(2  af  -  3(2  ay {3  6^)  +3(2  a)  (3  by  -  (3  b'y. 

This  shows  that  it  is  a  perfect  cube,  and  that  its  cube  root 
is  2  a  -  3  6^ 

Then,    8  a^  -  36  a'b^  +  54  ab'  -  27  6«  =  (2  a  -  3  b^. 

EXERCISE  16 

Factor  the  following : 

1.  x^  +  3x^  +  Sx  +  l. 

2.  8-12a  +  6a2_a3. 

3.  1  +  9  m  +  27  m2  +  27  m\ 

4.  64  n3  -  48  w2  +  12  w  -  1. 

5.  8  a3  +  36  a^6  +  54  a62  +  27  &3. 

6.  27  a^b^  -  108  a'^b^c  +  144  abc^  -  64  c^. 

7.  125  x^  -  600  a;2?/  +  960  xy^  -  512  yK 

8.  216  m6  +  756  m^x^  +  882  m2a;6  +  343  x^. 

177.  Case  IX.  When  the  expression  is  the  sum  or  difference 
of  two  perfect  cubes. 

By  §  138,  the  sum  or  difference  of  two  perfect  cubes  is 
divisible  by  the  sum  or  difference,  respectively,  of  their  cube 
roots. 

In  either  case  the  quotient  may  be  obtained  by  the  rules  of 
§138. 

1.  Factor  x^-21fz\ 

By  §  166,  the  cube  root  of  x^  is  a?,  and  of  27  2/V  is  3fz. 

Then  one  factor  is  a;^  —  3  'ifz. 

Dividing  x^  —  27  2/V  by  a.*^  —  3  ifz,  the  quotient  is 

a;^4-3ary2  +  9?/V  (§  138). 

Then,  x^  -  27  2/V  =  (a;^  -  3  fz)  {d"  +  3  ;x^fz  +  9  y^z"). 

2.  Factor  a^-\-b^. 
One  factor  is  a^  +  6^ 

Dividing  a^  +  W  by  a?  +  6^  the  quotient  is  a*  —  a^ft^  _|.  ^4^ 

Then,  a«  +  6«  =  (a^  +  ft^s^  {a'  -  a'b'  +  &*). 


90  ADVANCED   COURSE   IN   ALGEBRA 

3.    Factor  (x  +  af  —{x  —  of: 

{x  +  of  —  {x  —  ay 

=  \_{x  +  a)-{x-  a)]  [(a;  +  af  -\- {x  +  a)  (x  -  a) -\- {x  -  a)^] 
=  {x  -\-  a  —  X  -\-  a){x^  -\-2  ax  -\-  o?  -\-  x^  —  0?  -[■  x^  —  2  ax  -\-  a-) 
=  2a{3x'-\-a'). 

178.  Case  X.  Wheii  the  expression  is  the  sum  or  difference 
of  two  equal  odd  powers  of  two  numbers. 

By  §  142,  the  sum  or  difference  of  two  equal  odd  powers  of 
two  numbers  is  divisible  by  the  sum  or  difference,  respectively, 
of  the  numbers. 

The  quotient  may  be  obtained  by  laws  of  §  143. 

Ex.    Factor  a^  +  32  6^ 

By  §130,  32h'  =  {2hf. 

Then,  by  §  142,  one  factor  is  a  +  2  &. 
Dividing  a^  +  32  6^  by  a  +  2  h,  the  quotient  is 

a^  -  a^  (2  h)  +  a^  (2  hf  -  a  (2  hf  +  (2  by  (§  143). 

Whence, 

a*  +  32  6^=(a  +  2&)(a^-2a3&  +  4a262_8a&-^  +  16  60. 


Factor  the  following : 

EXERCISE  17 

1.    8w3-/i3.                 3. 

a^  +  64.                         5.    729  a%^  +612  c^^. 

2.    x^y''  +  125  z^.           4. 

216  a^m^  _  343  ,^9.       e.    TO3-(m  +  w)3. 

7.    {x-\-yy+{x-yY. 

9.   (2  a  +  x)3  -  (a  +  2  a;)3. 

8.    27(a-6)3-8  63. 

10.  .(5a;-2?/)3+(3x-4y)3. 

11.    x^  +  y^.                   14. 

1  +  a;7.                        17.   32  a^  -■  b^ 

12.    a^-\.                    15. 

m^  +  n^                      18,    2iSx^  +  y^ 

13.   a^-h\                  16. 

a^-1.                       19.   mi4  +  128n7. 

20.    1024  a565  _  243  cw. 

179.  By  application  of  the  rules  already  given,  an  expression 
may  often  be  resolved  into  more  than  two  factors. 

If  the  terms  of  the  expression  have  a  common  factor,  the 
method  of  §  155  should  always  be  applied  first. 


FACTORING  91 

1.  Factor  2  ax^y^  —  8  axy*. 

By  §  155,  2  ax^y-  -  8  axy'  =  2  axf  {y?  -^y") 

=  2axf{x-\-2y)(x-2y),hj^lll. 

2.  Factor  a«  -  h\ 

By  §171,  a^-h^={a^Jrh^)(p?-'b^- 

Whence,  by  §  177, 

a^-h^={a  +  h) (a? -ab  +  b') (a - b)(a'  +  ab-{-  6^. 

3.  Factor  af-y\ 

By  §  171,  x^-f=  {x'  +  y') {x' - y') 

=  (x'  +  y^){x^J^y^(x^-y^ 

=  {x*-\-y')(x'  +  y')(x  +  y){x-y). 

4.  Factor  3(m^ny-2(m'-n^). 

3(m-{-ny-2  (m^  -  n^)  =  3  (m -^  nf  -  2  (m -{-  7i)  (m  -  n) 
=  (m  +  n)[3  (m  +  n)  —  2  (m  —  n)] 
=  (m  +  n)(Sm-hSn  —  2m  +  2n) 
=  (m -\- n)  (m -{- 5  n). 

5.  Factor  a  (a- 1)- 6  (6-1). 

a(a-l)  -  b{b  -1)  =  a' -  a-b' -{-b 
=  a'-b'-a+b 
=  (a-\-b)(a-b)-{a-b) 
=  (a-b){a  +  b-l). 

EXERCISE  18 
Factor  the  following : 

1.  X^-  625.  10.  (16  m2  +  n^y  -  64  mH\ 

2.  ai2-l.  11.  2a'^x-Sa^x^  +  2a^z^-Sax\ 

3.  mi6  _  1 .  12.  9  a2c2  _  16  a2^2  _  36  ?,2c2 + 64  62(^2. 

4.  x^  -  26  a;3  -  27.  13.  x^*  -  2  x^  +  1. 

5.  (a-^  +  4  a6  +  62)2  _  (^2  +  ^2)2,      14.  729  -  n^ 

6.  12  ic6  -  18  xs  _  6  X*  +  9  x3.  15.  a^ft^  +  a'^yS  _  63a;2  _  3.2^^3. 

7.  81  TO*  -  256  ?i8.  16.  48  x^y  -  52  x^y^  -  140  xy^. 

8.  ai4  _  a;i4.  17.  16  a?  _  72  ae  +  108  ^5  _  54  a*.  ' 

9.  x6  -  16  x3i/3  +  64  ?/5,  18.  (rn  +  ny-2(m  +  ny+(m-\-ny. 


92     •  ADVANCED  COURSE   IN   ALGEBRA 

19.  Resolve  a^  +  512  into  three  factors  by  the  method  of  §  177. 

20.  a2  _  ,,^-2  ^a  +  m.  22.   ii^^  -  1024. 

21.  (x2  +  4  a;)2  _  37  (^2  +  4  ic)  + 160.    23.    m^  +  w  +  x^  +  x. 

24.  a2c2  _  4  52^2  _  9  ^2^2  +  35  52^2.  ^ 

25.  (m  -  n)  (x2  -  2/2)  +  (x  -f  y)  Qm"^  -  n^). 

26.  (X  -  1)3  4-  6(x  -  1)2  +  9(x  -  1). 

27.  a2_452_^_2&. 

28.  (m  +  n)  (m2  -  x2)  -  (m  +  x)  (m2  -  n^). 

29.  (x2  +  4  y2  _  ;22)2  _  16  x22/2.  31.    a^b^  +  27  a^y^  -  8  bH^ -216  x^yK 

30.  (x2-9x)2  +  4(x2-9x)-140.     32.    (2  x2  -  3)2  -  x2. 

33.    (m2  +  m)2  +  2(w2  +  m)  (w  +  1)  +  (w  +  l)^. 

34.  64a3x3  +  8a3_8x3- 1.  36.    (x +  2  i/)^  -  x(x2  -  4?/2). 

35.  (4  a2  _  62  _  9)2  _  35  52.  37.    (i  +  ^s^  +  (i  +  x)^ 

38.    (a2  +  6  a  +  8)2  -  14(^2  +  6  a  +  8)  -  15. 

39.  a4-9  +  2a(a2  +  3).  43.   m^  -  m^  +  32  m^  -  32. 

40.  (ix^-\-y^)-xy{x  +  y).  44.   a(a  -  c)- 6(&  -  c). 

41.  (a5_8m3)-«(a-2w)2.  45.   m%m  +  p)+ n'^(n -p). 

42.  18  a^d  +  22  a^b^  +  8  aft^.  46.   x^  +  8  x^  +  x^  +  8. 

47.  (27  m3  -  x3)  +  (3  w  +  x)  (9  m^  -  12  wx  +  x2). 

48.  (4  a2  +  9)2  -  24  a(4  a2  +  9)  4. 144  ^2. 

49.  m^  +  m^  -  64  m^  -  64. 

50.  (X2  +  ?/2)3  _  4  x22/2(x2  +  ?/2). 

51.  a5  4.  (^46  +  a^b-^  +  a263  _^  ^54  +  55. 

52.  (8w3-27)  +  (2w-3)(4w2  +  4w-6). 

180.   Factoring  by  Substitution. 

By  §  140,  if  the  expression  becomes  O'wlieii  x  is  put  equal  to 
a,  then  ic  —  a  is  a  factor. 

The  positive  and  negative  integral  factors  of  6  are  1,  2,  3,  6, 
-1,-2,-3,  and  -  6. 

It  is  best  to  try  the  numbers  in  their  order  of  absolute  mag- 
nitude. 

If  ic  =  1,  the  expression  becomes  1  —  7  -f  10  -|-  6. 

If  ic  =  —  1,  the  expression  becomes  —  1  —  7  —  10  -}-  6. 


FACTORING  93 

li  x  =  2,  the  expression  becomes  8  —  28  +  20  +  6. 

If  ic  =  —  2,  the  expression  becomes  —8  —  28  —  20  +  6. 

If  ic  =  3,  the  expression  becomes  27  —  63  +  30  +  6,  or  0. 

This  shows  that  re  —  3  is  a  factor. 

Dividing  the  expression  by  a;  —  3,  the  quotient  is  a^  —  4  a;  —  2. 

Then,  x^  -  7a^  +  10 a;  +  6  =  (a;  -  3)  (a;^  -  4a;  -  2). 

2.  Prove  that  a  is  a  factor  of 

(a-hb-\-c)  (ab  +  6c  +  cd)  -  (a  +  6)  (6  +  c)  (c  +  a). 
Putting  a  =  0,  the  expression  becomes 

(6  +  c)  6c  -  6  (&  +  c)  c,  orO. 
Then,  by  §  140,  a  is  a  factor  of  the  expression. 

3.  Prove  that  m  +  n  is  a  factor  of 

m*  —  4  m^n  +  2  m^n^  +  5  mn^  —  2  n*. 
Putting  m=  —rij  we  have 

7^4  ^  4^4  +  271^  -  5n*  -  2n*,  or  0. 
Then,  m  +  w  is  a  factor. 

EXERCISE  19 

Factor  the  following : 

1.  a;3  +  4xa  +  7x-12.  4.   «»  -  9a;2+ 15a;  + 9. 

2.  x4-x3  +  6x2  +  14a;  +  6.  5.   a;8-18a;  +  8. 

3.  x3  -  a;2  -  11 X  -  10.  6.   cc^  -  5x2  -  8x  +  4&  ; 

7.  a;4  +  83c3+13x2-13x-4. 

8.  2x4-7a;3  +  i0a;2-14a;  +  12. 

Find,  without  actual  division, 

9.  Whether  a;  -  3  is  a  factor  of  a;^  -  6  x^  +  13  x  -  12. 

10.  Whether  x  +  2  is  a  factor  of  x^  +  7  x2  _  6. 

11.  Whether  x  is  a  factor  of  x  (y  +  zy  +  y  {z  +  xy  -{-  z  (x -{■  yy. 

12.  Whether  a  is  a  factor  of  a^  (b  -  cy  -^  b^  (c  -  ay  +  c^  (a  -  by. 

13.  Whether  x  -  y  is  a  factor  of  (x  -  yy  +  (y  —  zy  +  {z  —  xy. 

14.  Whether  m  +  w  is  a  factor  of  w  (m  +  2n)8  —  n  (2  w  +  ny. 

15.  Whether  a  +  &  +  c  is  a  factor  of 

a  (6  +  c)  +  6  (c  4-  a)  +  c  (a  +  6)  +  a2  +  62  _f.  c2. 


94  ADVANCED   COURSE  IN  ALGEBRA 

181.   Factoring  of  Symmetrical  Expressions. 

The  method  of  §  180  is  advantageous  in  the  factoring  of  sym- 
metrical expressions.     (§§  146,  147.) 

1.  Factor 

a{b  +  cY  +  h{G->ray  +  c{a->thy-a\h  +  c)-h\c  +  a)-c\a  +  h). 

The  expression  is  symmetrical  with  respect  to  a,  6,  and  c. 

Being  of  the  third  degree,  the  only  literal  factors  which  it 
can  have  are  three  of  the  type  a ;  three  of  the  type  a  +  6 ;  or 
a  +  h  -\-  c,  and  a  factor  of  the  second  degree. 

Putting  a  =  0,  the  expression  becomes 

hd'-^ch^-Wc-c'h,  orO. 

Then,  by  §  140,  a  is  a  factor ;  and,  by  symmetry,  h  and  c 
are  factors. 

The  expression,  being  of  the  third  degree,  can  have  no  other 
literal  factor ;  but  it  may  have  a  numerical  factor. 

Let  the  given  expression  =  mdbc. 

To  determine  m,  let  a  =  6  =  c  =  1. 

Then,  4 +  4 +4-2-2- 2  =  m,  or  m  =  6. 

Whence,  the  given  expression  =  6  dbc. 

2.  Factor  x\y  -\- z) -{- y"^  (z -\- x)  +  z^  {x -\- y)  +  ^  xyz. 

The  expression  is  symmetrical  with  respect  to  x,  y,  and  z. 

The  only  literal  factors  which  it  can  have  are  three  of  the 
type  X ;  three  of  the  type  a;  +  ?/  j  ot  x  -\-  y  -{-  z,  and  a  factor  of 
the  second  degree. 

It  is  evident  that  neither  x,  y,  nor  2;  is  a  factor. 

Putting  X  equal  to  —  y,  the  expression  becomes 

y%y-\-z)-{-y\z-y)-Sy% 

which  is  not  0. 

Then,  x-{-y  is  not  a  factor ;  and,  by  symmetry,  neither  y-\-z 
nor  2;  +  a;  is  a  factor. 

Putting  X  equal  to  —y  —  z,  the  expression  becomes 
(y  +  ^)\y  ■^z)-f-z^-3  yz{y  +  z) 

=  y^  -{-^  y'z  +  3  yz^ -{- z^ -  f  -  z^ -S  fz  -Syz^  =  0. 

Then,  a;  +  2/  +  2;  is  a  factor. 


FACTORING  ,        95 

The  other  factor  must  be  of  the  second  degree ;  and,  as  it  is 
symmetrical  with  respect  to  x,  y,  and  z,  it  must  be  of  the  form 
m  {x^  +  2/^  +  2!^),  or  ii  (xy  -{-  yz  +  zx). 

The  first  of  these  cannot  be  a  factor ;  for,  if  it  were,  there 
woukl  be  terms  involving  x^,  y^,  and  z^  in  the  given  expression. 

Then,  the  given  expression  =  7i(x  -{-y  -{-  z)(xy  -{-yz-\-  zx). 

To  determine  n,  let  x  =  1,  y  =  1,  and  z  =  0. 

Then,  1  +  1  =  2n,  and  n  =  l. 

Then,  the  given  expression  =(x-\-  y  -^z)  (xy  -\-yz-{-  zx). 

3.    Factor  ab  (a  —  6)  +  be  (6  —  c)  +  ca  (c  —  a). 

The  expression  is  cyclo-symmetrical  with  respect  to  a,  6, 
and  c. 

It  is  evident  that  neither  a,  b,  nor  c  is  a  factor. 

The  expression  becomes  0  when  a  is  put  equal  to  b. 

Then,  a  —  b  is  a  factor;  and,  by  symmetry,  b  —  c  and  c  —  a 
are  factors. 

The  expression  can  have  no  other  literal  factor,  but  may 
have  a  numerical  one. 

Let  the  given  expression  =m(a—  b)(b  —  c)(c  —  a). 

To  determine  m,  let  a  =  2,  6  =  1,  and  c  =  0. 

Then,  2  =  —  2  m,  and  m  =  —  1. 

Then,  the  given  expression  =  —(a  —  b) (b  —  c) (c  —  a). 

EXERCISE  20 

Factor  the  following : 

1.  m3  +  2  TO%  +  2  wn2  +  n^ 

2.  (ab  +  &c  +  ca)(a  +  6  +  c)  -  a^  (6  +  c)  -  b^  (c  +  a)  -  c^  (a  +  6). 

3.  x2  (?/  +  ^)  +  yi  (^z-i-x)  +  z^(x  +  y)  +  2  xyz. 

4.  a  (ft  +  c)2  +  6  (c  +  a)2  +  c  (a  +  &)2  -  4  a6c. 

5.  aP-  (6  -  c)  +  &2  (c  _  a)  +  c2  (a  -  6). 

6.  {x-\-y  +  z)  (xy  +  yz  +  zx)  -  (x  +  y)  (y  +  z)  (z  +  «). 

7.  ab  (a  +  &)  +  6c  (&  +  c)  +  ca  (c  +  a)  +  2  a6c. 

8.  (X  +  y  +  0)3  -  (x3  +  y^  +  z^). 

9.  (x  +  y  +  0) (xy  +  yz  +zx)  -  xyz. 

10.  (x  -  yy  +  (y-  zy  +  (z-  x)3. 

11.  a3  (6  -  c)  +  63  (c  -  a)  +  c^  (a  -  6). 


96  ADVANCED   COURSE   IN   ALGEBRA 

SOLUTION  OF  EQUATIONS  BY  FACTORING 

182.   Consider  the  equation 

AxBxGx-"  =  0;  (1) 

where  Aj  B,  C,  •••,  are  integral  expressions  which  involve  the 
unknown  numbers. 

By  §  49,  if  Ax  B  xG  X  '"  =0,  some  one  of  the  factors  A, 
B,  C,  •••,  must  equal  0. 

We  obtain  in  this  way  a  series  of  equations 

A=0,B  =  0,G  =  0,'".  (2) 

We  will  now  show  that  these  are  equivalent  (§  114)  to'(l). 

Any  solution  of  (1)  makes  Ax  B  x  G  X  •  •  •  identically  equal 
toO. 

It  then  makes  at  least  one  of  the  factors  A,  B,  G,  •••,  identi- 
cally equal  to  0 ;  and  hence  satisfies  at  least  one  of  the  equa- 
tions (2). 

Again,  any  solution  of  any  of  the  equations  (2)  makes  A  X 
B  X  G  X  ••'  identically  equal  to  0;  and  hence  satisfies  (1). 

Then,  (1)  and  (2)  are  equivalent. 

It  follows  from  the  above  that  the  equation 
AxBxGX"'=(i 
may  be  solved  by  placing  the  factors  of  the  first  member  sepa- 
rately equal  to  zero,  and  solving  the  resulting  equations. 

1.  Solve  the  equation  2x^  —  x  =  0. 

Factoring  the  first  member,  the  equation  becomes 

aj(2a;-l)  =  0  (§155). 
Placing  the  factors  separately  equal  to  0, 
«  =  0; 
and  2  aj  —  1  =  0,  or  a;  =  -• 

2.  Solve  the  equation  ar^-f-4x-  —  ic  —  4  =  0. 
Factoring  the  first  member  (§§  156,  171), 

(a;  +  4)(a^-l)  =  0,  or  (a;  +  4)  (a; -f  1)  (a;  - 1)  =  0. 


FACTORING  97 

Then,  a;  +  4  =  0,  orx=— 4; 

aj  + 1  =  0,  ov  x=  —  1 ; 
and  a;  —  1  =  0,  or  X  =  1. 

3.   Solve  the  equation  (2x-  3f  =(x-  ly  -\-{x-  2y. 
Factoring  the  second  member,  we  have 
[{x-l)  +  {x-2)^l{x-iy-(x-l)(x-2)  +  (x-2y] 
=  (2x-S)(x'-2x-\-l-x'-{-Sx-2  +  x'-4:X-\-4:) 
=  (2x-3)(x'-3x  +  3). 
Then  the  given  equation  can  be  written 

(2a; -3)  [(2a;- 3)2- (a^ -3a;  +  3)]  =  0. 
Or,  (2a;-3)(4a;2-12a;4-9-a;2-f-3a;-3)  =  0. 
Or,  (2a;-3)(3a;2-9a;  +  6)  =  0. 

Dividing  both  members  by  3, 

(2a;-3)(a;2-3a;  +  2)  =  0,  or  (2  a;- 3)  (a;- 1)  (a;- 2)  =  0. 
Then,  2a;-3  =  0,  or  a;  =  ?; 

a;  — 1  =  0,  ora;  =  l; 

and  a;  —  2  =  0,  or  a;  =  2. 

The  above  examples  illustrate  the  principle  (§  715)  that  the  degree 
(§  113)  of  an  equation  involving  one  unknown  number  indicates  the  num- 
ber of  its  roots  ;  thus,  an  equation  of  the  third  degree  has  three  roots  ;  of 
the  fourth  degree,  four  roots;  etc.  It  should  be  observed  that  the  roots 
are  not  necessarily  unequal ;  thus,  the  equation  x^  —  2x  +  l  =  0  may  be 
written  (x  —  1)  (a;  —  1)  =  0,  and  therefore  the  two  roots  are  1  and  1. 

EXERCISE  21 

Solve  the  following  equations : 

1.  5a;4  +  35a;3  =  0.  8.  ic* -  18 a;8  +  32 ^2  =  0. 

2.  3x3- 108a;  =  0.  9.  6x2  + 7a; +  2  =  0. 

3.  (4x-3)(4x2-25)=0.  10.  10x2  -  7x- 12  =  0. 

4.  x2  +  23 x  + 102  =  0.  11.  15x2 +  x -2  =  0. 

5.  x2  +  4x-96  =  0.  12.  12x3-29x2  +  15x  =  0. 

6.  x2  -  17  X- 110  =  0.  13.  x2-ax  +  &x-a6  =  0. 

7.  (5x  +  l)(x2  +  22x.+  121)=0.  14.  x2  + mx  + 7ix  + mn  =  0. 


98  ADVANCED   COURSE   IN   ALGEBRA 

15.  x^-2cx-Sx-\-lQc  =  0.  16.    x^  +  Sm'^x-bin^x- I5m^=0. 

17.  (4x2-28x+49)(ic2-3x-10)(8ic2+  14X-15)  =0. 

18.  27x3  +  18x'-3x-2  =0. 

19.  a;3  +  6x2-ic-30zr:0  (§180). 

20.  x4  +  2x3-13x2-14x+24  =  0. 

21.  (x-2)2-4(x-2)+3  =  0.  22.    (x  -  2)3  +  8x3=  (8x- 2)3. 

23.  x2-4-(x-2)(3x2  +  4x-4)=0. 

24.  (x2-l)(x2-9)= -3(x-l)(x  +  3). 

25.  (2x+l)3- (x+2)3=(x-l)3. 

26.  (x2-l)(x3-8)  -19(x  +  l)(x2-3x  +  2)  =0. 

•     183.   It  follows  from  §§  140  and  182  that 
If  the  first  member  of  the  equation 

A  =  0 

is  a  rational   and   integral  polynomial  involving  the  unknown 
number  x,  and  divisible  by  x  —  a,  the  equation  has  a  as  a  root. 

For  by  §  140,  the  first  member  has  x  —  a  as  a  factor. 

If  A  is  divisible  by  ax  —  6,  the  equation  has  -  as  a  root. 

a 

2 

Ex.    Find  whether is  a  root  of  the  equation 

o 

3x'-{-Sa^-^13x'  +  9x-i-2  =  0. 

Dividing  the  first  member  by  3  .x  -f-  2,  the  quotient  is  aP-\-2x'^ 
2 

-f  3  a?  -f- 1 ;  then, is  a  root. 

o 

EXERCISE  22 

Find  whether : 

1.  4  is  a  root  of  x3  -  x2  -  19  x  +  28  =  0. 

2.  -  I  is  a  root  of  6x3+ 13x2  + 5x  + 25  =  0. 

3.  -  is  a  root  of  4  x3  -  11  x2  -  14  x  -  15  =  0. 
4 

4.  -  5  is  a  root  of  4  x*  +  22  x^  +  9  x2  -  8  x  -  15  =  0. 

5.  -  is  a  root  of  15 x*  -  17x3  +  7x2-  19x+  6  =  0. 

6.  -  3  is  a  root  of  9x*  +  26x3  -  8x2  -  11  x  -  3  =  0. 


HIGHEST  COMMON   FACTOR  99 


IX.   HIGHEST  COMMON  FACTOR.     LOWEST 
COMMON  MULTIPLE 

HIGHEST  COMMON  FACTOR 

In  the  present  chapter,  we  consider  only  rational  and  integral  expres- 
sions (§  63),  with  integral  numerical  coefficients. 

184.  The  Highest  Common  Factor  (H.  C.  F.)  of  two  or  more 
expressions  is  their  common  factor  of  highest  degree  (§  64) ; 
or  if  several  common  factors  are  of  equally  high  degree,  it  is 
the  one  having  the  numerical  coefficient  of  greatest  absolute 
value  in  its  term  of  highest  degree. 

There  are  always  two  forms  of  the  highest  common  factor,  one  of 
which  is  the  negative  of  the  other. 

Thus,  in  the  expressions  a^  —  ah  and  6^  _  q,^^  either  a  —  h  or  h  —  a 
will  exactly  divide  each  expression. 

185.  Two  expressions  are  said  to  be  prime  to  each  otKer 
when  unity  is  their  highest  common  factor. 

In  determining  the  highest  common  factor  of  expressions,  it 
is  convenient  to  distinguish  two  cases. 

186.  Case  I.  When  the  expressions  are  monomials,  or  poly- 
nomials ivhich  can  be  readily  fax^tored  by  inspection. 

1.  Required  the  H.  C.  F.  of  42  a^H',  70  a^bc,  and  98  a'bH\ 
The  H.  C.  F.  of  42,  70,  and  98^s  14. 

It  is  evident  1)y  inspection  that  the  expression  of  highest 
degree  which  will  exactly  divide  a^b'\  d^bc,  ^nd  a*b^d^,  is  a^b. 

Then,  the  H.  C.  F.  of  the  given  expressions  is  14  a^b. 

It  will  be  observed,  in  the  above  result,  that  the  exponent  of 
each  letter  is  the  loiuest  exponent  with  which  it  occurs  in  any  of  the 
given  expressions. 

2.  Required  the  H.  C.  F.  of 

5  x'y  -  45  x'y  and  10  xh/  -f-  40  xhf  -  210  xy\ 


100  ADVAN^CED  COURSE   IN  ALGEBRA 

By  §§  155,  171,  and  173, 

5  x'^y  —  45  x^y  z=5a^y  (x^  —  9) 

=  5x'y{x-]-S)(x-3), 
and  10  x^y^  +  ^Ox^y^  -  210xy^  =  10 xy^ (x" -{- 4. x  -  21) 
,         =10xy'(x-{-7)(x-S). 
The  H.  C.  F.  of  the  numerical  coefficients  5  and  10  is  5. 
It  is  evident  by  inspection  that  the  H.  C.  F.  of  the  literal 
portions  of  the  expressions  is  xy  (x  —  3). 

Then,  the  H.  C.  F.  of  the  given  expressions  is  5xy(x  —  3). 

EXERCISE  23       . 

Find  the  highest  common  factor  of  : 

1.  64  x3  +  27?/3,  16  ic2  -  9  y^,  and  16  x^-\-2ixy  +  9  y'^. 

2.  2x^-  12  x2  +  16  x,  Sx^-Sx^-  36  a;2,  and  6x^  +  bx'^-  100  x\ 

3.  125  w3  -  8,  10  w2  +  m  -  2,  and  25  m2  -  20  m  +  4. 

4.  a*  -  3  a2  _  28,  a*  -  16,  and  a^  +  a'-j  ^-  4  a  +  4. 

5.  2  ic3  ^  ic2  _  6  X  -  3,  6  a;2  +  19  a  +  8,  and  8  x^  +  12  x2  +  6  a;  +  1. 

6.  27  x3  -  ?/8,  243  x^  -  y^,  and  12  x2  -  25  a;?/  +  7  ?/2. 

7.  a2  +  &2  +  c2  -  2  a6  4-  2  ac  -  2  6c,  a2  +  &2  _  c2  _  2  ah,    and 
flj2  _  52  _  c2  _|-  2  6c, 

8.  27  a3  +  135  a26  +  225  ah^  +  125  6^,  3  a2  +  2  a6  -  5  62,    and 
3  ac  -  6  ad  +  5  6c  -  10  hd, 

9.  4x*  +  11  a;2  +  25,  2  x^  -  9  a;2  +  14  x  -  15,  and  2  x^  +  x2  -  x  +  10. 

187.  Case  II.  When  the  expressions  are  polynomials  which 
cannot  he  readily  factored  hy  inspectiori. 

Let  A  and  B  be  two  polynomials,  arranged  according  to  the 
descending  powers  of  some  common  letter,  and  let  the  exponent 
of  that  letter  in  the  first  term  of  A  be  not  lower  than  its  ex- 
ponent in  the  first  term  of  B. 

Suppose  that,  when  A  is  divided  by  B,  the  quotient  is  p,  and 
the  remainder  C. 

We  will  prove  that  the  H.  C.  F.  of  B  and  C  is  the  same  as 
the  H.  C.  F.  of  A  and  B. 

The  operation  of  division  is  shown  as  follows  : 


HIGHEST   COMMON   FACTOR  101 

B)A(p 
pB 

~d~ 

We  will  first  prove  that  every  common  factor  of  B  and  G  is 
a  common  factor  of  A  and  B. 

Let  F  be  any  common  factor  of  B  and  C ;  and  let 

B=bF,  and  C=cF.  (1) 

Since  the  dividend  is  equal  to  the  product  of  the  quotient 
and  the  divisor,  plus  the  remainder,  we  have 

A=pB  +  a  (2) 

Substituting  in  (2)  the  values  of  B  and  C  from  (1), 

A=pbF-{-cF=F(pb-\-c).  (3) 

It  is  evident  from  (1)  and  (3)  that  i^  is  a  common  factor  of 
A  and  B. 

We  will  next  prove  that  every  common  factor  of  A  and  B  is 
a  common  factor  of  jB  and  C. 

Let  F'  be  any  common  factor  of  A  and  B  ;  and  let 

A  =  mF',  and  B  =  nF',  (4) 

From  (2),  C=A-pB 

—  mjri  —  pnF'  =  F'  (m—  pn).  (5) 

.  From  (4)  and  (5),  F'  is  a  common  factor  of  B  and  C. 

It  follows  from  the  above  that  the  H.  C.  F.  of  5  and  C  is 
the  same  as  the  H.  C.  F.  of  A  and  B. 

188.   Let  A,  B,  and  C  have  the  same  meanings  as  in  §  187. 

Suppose  that  when  B  is  divided  by  (7,  the  quotient  is  q,  and 
the  remainder  D ;  that  when  C  is  divided  by  D,  the  quotient  is 
r,  and  the  remainder  E,  and  so  on ;  and  that  we  finally  arrive 
at  a  remainder  H,  which  exactly  divides  the  preceding  divi- 
sor G. 

By  §  187,  the  H.  C.  F.  of  C  and  D  is  the  same  as  the  H.  C.  F. 
of  B  and  C;  the  H.  C.  F.  of  D  and  E  is  the  same  as  the  H.  C.  F. 
of  C  and  D ;  and  so  on. 


102  ADVANCED   COURSE  IN   ALGEBRA 

Hence  the  H.  C.  F.  of  G  and  H  is  the  same  as  the  H.  C.  F. 
of  A  and  B. 

But  since  ^exactly  divides  O,  His  itself  the  H.  C.  F.  of  G 
and  H. 

Therefore,  H  is  the  H.  C.  F.  of  A  and  B. 

We  derive  from  the  above  the  following  rule  for  the  H.  C.  F. 
of  two  polynomials,  A  and  B,  arranged  according  to  the  de- 
scending powers  of  some  common  letter,  the  exponent  of  that 
letter  in  the  first  term  of  A  being  not  lower  than  its  exponent 
in  the  first  term  of  B : 

Divide  A  by  B. 

If  there  be  a  remainder,  divide  the  divisor  by  it;  and  continue 
thus  to  make  the  remaiyider  the  divisor,  and  the  preceding  divisor 
the  divideyid,  until  there  is  no  remainder. 

The  last  divisor  is  the  H.  C.  F.  required. 

Note  1.  It  is  important  to  keep  the  work  throughout  in  descending 
powers  of  some  common  letter ;  and  each  division  should  be  continued 
until  the  exponent  of  this  letter  in  the  first  term  of  the  remainder  is  less 
than  its  exponent  in  the  first  term  of  the  divisor. 

Note  2.  If  the  terms  of  one  of  the  expressions  have  a  common  fac- 
tor which  is  not  a  common  factor  of  the  terms  of  the  other,  it  may  be 
removed  ;  for  it  can  evidently  form  no  part  of  the  highest  common  factor. 
In  like  manner,  we  may  divide  any  remainder  by  a  factor  which  is  not  a 
factor  of  the  preceding  divisor. 

Note  3.  If  the  first  term  of  the  dividend,  or  of  any  remainder,  is  not 
divisible  by  the  first  term  of  the  divisor,  it  may  be  made  so  by  multiply- 
ing the  dividend  or  remainder  by  any  term  which  is  not  a  factor  of  the 
divisor. 

Note  4.  If  the  first  term  of  any  remainder  is  negative,  the  sign  of 
each  term  of  the  remainder  may  be  changed.     (See  note  to  §  184.) 

Note  5.  If  the  given  expressions  have  a  common  factor  which  can 
be  seen  by  inspection,  remove  it,  and  find  the  H.  C.  F.  of  the  resulting 
expressions. 

The  result,  multiplied  by  the  common  factor,  will  be  the  H.  C.  F.  of 
the  given  expressions. 

Note  6.  The  operation  of  division  may  usually  be  abridged  by  thej 
use  of  detached  coefficients  (§  104). 


HIGHEST  COMMON  FACTOR  103 

1.  Kequired  the  H.C.F.  of 

6a^-{-7a^b-Sab^  and  4.a^b  +  S  d'b^ -Sa¥-^b\ 

We  remove  the  factor  a  from  the  first  expression,  and  the 
factor  b  from  the  second  (Note  2). 
We  then  find  the  H.  C.  F.  of 

6a^  +  7ab-3b^  and  4:a^-\-S  a'b-S  ab^-9  b^ 

Since  4  a^  is  not  divisible  by  6  a^,  we  multiply  the  second 
expression  by  3  (Note  3). 

4.a^  +  Sa'b-3ab^-9b^ 
3 

6 a^  +  7  ab -3  b^)12  a^  +  24.  a^b -9  ab^ -27  b\2  a 
12a^-{-Ua'b-6ab^ 

10  a'b- 3  ab^-27b^ 

Since  10  a^b  is  not  divisible  by  6  a^,  we  multiply  this  remain- 
der by  3. 

60.^+7  ab-3b^30a'b-   9aW-Slb\6b 
30  o?b  +  35  ab^  - 15  W 
~         r44a62-66  6« 
Dividing  the  remainder  by  —  22  b^  (Notes  2,  4), 
2  a  +  3  6)6  a2  -f-  7  a6  -  362(3a  -  6 

-2a6' 
-2ab-3b^ 


Then,  2  a  +  3  5  is  the  H.  C.  F.  required. 

2.   Kequired  the  H.  C.F.  of 

2x^-3^  -x^  +  x   and   6a;*  -  a^ +  3ar^  -  2a;. 

E-emoving  the  common  factor  x  (Note  5),  and  using  De- 
tached Coefficients, 

2-3-1  +  1)6-    1+    3-    2(3* 
6-.  9-    3+    3 
8+    6-    5 


104  ADVANCED  COURSE  IN  ALGEBRA 


2-    3-    1+    1 
4 

8  +  6- 

_  5)8 -12-    4+    4(1 
8+    6-    5 

-18+    1+    4 

4 

-72+    4  +  16(-9 

-72-54  +  45 

29)58  -  29 

2-    1 

2-1)8+    6-    5(4  +  5 
8-    4 

10-5 
10-5 

The  last  divisor  is  2x  —  l. 

Multiplying  this  by  x,  the  H.  C.  F.  of  the  given  expressions 

is  ic(2aj  — 1). 

189.  The  H.  C.  F.  of  three  expressions  may  be  found  as 
follows : 

Let  A,  B,  and  C  be  the  expressions. 

Let  G  be  the  H.  C.  F.  of  A  and  JB ;  then,  every  common 
factor  of  G  and  O  is  a  common  factor  of  A,  B,  and  C. 

But  since  every  common  factor  of  two  expressions  exactly 
divides  their  H.  C.  F.  (§  188),  every  common  factor  of  A,  B, 
and  C  is  also  a  common  factor  of  G  and  C. 

Whence,  the  H.  C.  F.  of  G  and  C  is  the  H.  C.  F.  of  A,  B, 
and  C. 

Hence,  to  find  the  H.  C.F.  of  three  exp7*essions,  find  the  H.  C.  F. 
of  two  of  them,  and  then  of  this  result  and  the  third  expression. 

We  proceed  in  a  similar  manner  to  find  the  H.  C.  F.  of  any 
number  of  expressions. 

EXERCISE  24 

Find  the  highest  common  factor  of  ; 

1.   4  a;2  +  4  x  -  3  and  6  x^  +  11  a;2  -  a;  -  6. 


LOWEST   COMMON  MULTIPLE  105 

2.  6a^-  lla^b  -7ab^  +  4  fes  and  12  a^  -  13  a'^b  +  21  ab^-6b^ 

3.  9a;4-21x3  +  48a;2_24x  and  ISx*  -  25a;3  +  25x2  -  55  a;  +  30. 

4.  6  a*  +  a8  +  5  a2  +  7  a  _  3  and  8  a'^  -  6  a^  +  7  a^  -  9. 

5.  6x5  +  a:4  +  3a;3-6x2-4a:  and  12x5 +8a;4  -  3x3  -  lOx^ -4  a;. 

6.  8x2-6x-35,  10x3-27 X2-X  +  15,  and  6x3-13x2-13x4- 20. 

7.  6  a3  _  19  a%  +  a62  +  6  63,    8  a3  -  18  a'^b  -  17  a&2  _  3  63,   and 
6  a3  +  23  a^b  -  6  ab'^  -  S  b^ 

LOWEST  COMMON  MULTIPLE 

190.  A  Common  Multiple  of  two  or  more  expressions  is  an 
expression  which,  is  exactly  divisible  by  each  of  them. 

191.  The  Lowest  Common  Multiple  (L.  C.  M.)  of  two  or  more 
expressions  is  their  common  multiple  of  lowest  degree;  or  if 
several  common  multiples  are  of  equally  low  degree,  it  is  the 
one  having  the' numerical  coefficient  of  least  absolute  value  in 
its  term  of  highest  degree. 

There  are  always  two  forms  of  the  lowest  common  multiple,  one  of 
which  is  the  negative  of  the  other;  thus,  in  the  expressions  a^  —  ab 
and  62  _  ^55,  either  ab(a  —  6)  or  a6(6  —  a)  is  exactly  divisible  by  each 
expression. 

In  determining  the  lowest  common  multiple  of  expressions, 
it  is  convenient  to  distinguish  two  cases. 

192.  Case  I.  When  the  expressions  are  monomials^  or  poly- 
nomials ivhich  can  be  readily  factored  by  inspection. 

1.  Required  the  L.  C.  M.  of  36  A*,  60  ay,  and  84ca;^ 
The  L.  C.  M.  of  36,  60,  and  84  is  1260. 

It  is  evident  by  inspection  that  the  expression  of  lowest 
degree  which  is  exactly  divisible  by  a^x,  a^y-,  and  ca^,  is  a^cx^y^. 

Then,  the  L.  C.  M.  of  the  given  expressions  is  1260  a^cx^y^. 

It  will  be  observed,  in  the  above  result,  that  the  exponent  of 
each  letter  is  the  highest  exponent  with  ivhich  it  occurs  in  any  of 
the  given  expressions. 

2.  Required  the  L.  C.  M.  of 

a;^  -f  a;  —  6,  a^  —  4  a;  -|-  4,  and  a^  —  9  a;. 


106  ADVANCED  COURSE   IN  ALGEBRA 

By  §  173,  x^  +  x-6  =  (x  +  'S){x-2). 

By  §  169,  x'-4.x  +  4.  =  (x-2y. 

By  §  171,  ix^-9x  =  x{x  +  3){x-S). 

It  is  evident  by  inspection  that  the  L.  C.  M.  of  these  expres- 
sions is  x(x  —  2y  (x  +  3)  (a;  —  3). 

EXERCISE  25 

Find  the  lowest  common  multiple  of : 

1.  x^-16x  +  b0,x^-{-2x-  35,  and  x2  -  3 a;  -  70. 

2.  27  a*  +  64  a,  18  a*  -  32  a"-,  and  3  a2  +  7  a  +  4. 

3.  6  ic2  +  7  a;  -  5,  10  x2  -  9  ic  +  2,  and  8x^  -  12x'^ -{- 6x -1. 

4.  Sac+  ad-6bc-2bd,ac-4ad-2bc  +  S  bd,  and  3  c2  -  11  cd-Ad^. 
6.  a2  +  4  62  _  9  c2  _  4  a&,  a2  _  4  52  _  9  c2  +  12  be,  and 

^2  _{.  4  52  _|_  9  c2  _  4  Q5&  -  6  ac  +  12  6c. 

6.  8  TO»  -  w3,  4  m2  -  4  mw  +  w2,  and  16  m*  +  4  w2?i2  +  n*. 

7.  a;2  +  5 X  +  6,  x3  -  19 X  -  30,  and  x^-7x^  +  2x  +  40. 

193.  Case  II.      Wheyi  the  expressions  are  polynomials  which 
cannot  be  readily  factored  by  inspection. 

Let  A  and  B  be  any  two  expressions. 

Let  F  be  their  H.  C.  F.,  and  M  their  L.  C.  M. ;   and  suppose 
that  A  =  aF,  and  B  =  bF. 

Then,  AxB  =  abF\  (1) 

Since  F  is  the  H.  C.  F.  of  A  and  5,  a  and  &  have  no  common 
factors ;  whence,  the  L.  C.  M.  of  aF  and  bF  is  abF. 

That  is,  M=  abF. 

Multiplying  each  of  these  equals  by  F,  we  have 

FxM=abF\  (2) 

From  (1)  and  (2),       A  x  B  =  F  x  M.  (Ax.  4,  §  66) 

That  is,  the  product  of  two  expressions  is  equal  to  the  product 
of  their  H.  C.  F.  and  L.  C.  M. 

194.  It  follows  from  §  193  that,  to  find  the  L.  C.  M.  of  two 
expressions. 

Divide  their  product  by  their  highest  common  factor. 


LOWEST   COMMON  MULTIPLE  107 

Or,  divide  one  of  the  expressions  by  their  highest  common  factor, 
and  multiply  the  quotient  by  the  other  expression. 

Ex,     Required  the  L.  C.  M.  of 

x^  -Sxy  -^7  y^  Sind  a^  -9  afy  +  23  xy^  -  15  f. 

By  the  rule  of  §  188,  the  H.  C.  F.  of  the  given  expressions  is 
x-y. 

Dividing  x^  —  S  xy  -{-  7  y^  hj  x  —  y,  the  quotient  is  x—7  y. 
Then,  the  L.  C.  M.  of  the  given  expressions  is 
(x-7y){a^-9a^y-h2S  xy^  -15f). 

195.  It  follows  from  §  193  that,  if  two  expressions  are  prime 
to  each  other  (§  185),  their  product  is  their  L.  C.  M. 

196.  The  L.  G.  M.  of  three  expressions  may  be  found  as 
follows : 

Let  A,  B,  and  C  be  the  expressions. 

Let  M  be  the  L.  C.  M.  of  A  and  B ;  then  every  common  mul- 
tiple of  M  and  (7  is  a  common  multiple  of  A,  B,  and  C. 

But  since  every  common  multiple  of  two  expressions  is  ex- 
actly divisible  by  their  L.  C.  M.,  every  common  multiple  of  A, 
B,  and  C  is  also  a  common  multiple  of  M  and  C. 

Whence,  the  L.  C.  M.  of  M  and  C  is  the  L.  C.  M.  of  A,  B, 
and  C. 

Hence,  to  find  the  L.  C.  M.  of  three  expressions,  find  the  L.  C.  M. 
of  two  of  them,  and  then  of  this  result  and  the  third  expression. 

We  proceed  in  a  similar  manner  to  find  the  L.  C.  M.  of  any 
number  of  expressions. 

EXERCISE  26 

Find  the  lowest  common  multiple  of  : 

1.  8  ic2  -  6  X  -  9  and  6  ic8  -  7  ic2  -  7  X  +  6. 

2.  6  a8  +  a^h  -  11  a&2  -  6  68  and  6  «»  -  5  a^b  -^ab-  +  Z  b\ 

3.  8  x8  -  22  a;2  _  6  X  and  8  a:6  ^  6  a;6  _  11  x*  -  23  x8  -  5  xK 

4.  X*  -  2  x8  -  2  x2  +  7  X  -  6  and  x*  -  4  x^  +  x^  +  7  x  -  2. 

5.  4  a8  +  4  a2  _  43  <x  +  20,   4  a^  +  20  a^  +  13  «  _  12,    and 
4  a3  +  12  a2  -  31  a  -  60. 

6.  6  x2  -  7  X  -  3,  4  x3  -  4  x2  4-  3  X  -  9,  and  4  x3  -  12  x2  -  X  +  15. 


108  ADVANCED   COURSE  IN   ALGEBRA 


X.    FRACTIONS 

197.  In  the  fraction  ^,  the  dividend  a  is  called  the  numera- 

h 

tor,  and  the  divisor  h  the  denominator. 

The  symbol  /  is  often  used  to  represent  a  fraction ;  thus,  a/h  signifies  -• 

h 

The  numerator  and  denominator  are  called  the  terms  of  the 

fraction. 

198.  A  rational  fraction  is  a  fraction  whose  terms  are  rational 
and  integral  (§  63). 

A  monomial  is  said  to  be  rational  wheu  it  is  a  rational  and 
integral  expression,  or  a  rational  fraction.  • 

A  polynomial  is  said  to  be  rational  when  each  of  its  terms  is 
rational. 

199.  By  §96,(1),  ?  =  ^. 

That  is,  if  the  terms  of  a  fraction  he  both  multiplied  by  the  same 
expression,  the  value  of  the  fraction  is  not  changed. 

a 

200.  By  §  30,  (2),  |  =  |3«  =  f- 

c 

That  is,  if  the  terms  of  a  fraction  be  both  divided  by  the  same 
expression,  the  value  of  the  fraction  is  not  changed. 

201.  By  §48,      ±±^ZZ^  =  -±±^-^^. 

^  "^      '      -\-b      -b  -b  +6 

That  is,  if  the  signs  of  both  terms  of  a  fraction  be  changed,  the 
sign  before  the  fraction  is  not  changed;  but  if  the  sign  of  either 
one  be  changed,  the  sign  before  the  fraction  is  changed. 

If  either  term  is  a  polynomial,  care  must  be  taken,  on  chang- 
ing its  sign,  to  change  the  sign  of  each  of  its  terms. 


FRACTIONS  109 

Thus,  the   fraction       ~  ,  by  changing  the  signs  of  both 
c  —  d  J  _ 

numerator  and  denominator,  can  be  written  (§  81). 

d  —  c 

202.   It  follows  from  §§91  and  201  that 

If  either  term  of  a  fraction  is  the  indicated  product  of  two  or 
more  expressions,  the  signs  of  any  even  number  of  them  may  he 
changed  ivithout  changing  the  sign  before  the  fraction;  but  if  the 
signs  of  any  odd  number  of  them  be  changed,  the  sign  before  the 
fraction  is  changed. 

Thus,  the  fraction ^ may  be  written 

^c-d)(e-f) 


a  —  b  b  —  a 


etc. 


(d-c)(f-e)     (d-c){e-f)         id-c){f-e) 

REDUCTION  OF  FRACTIONS  • 

203.  Reduction  of  a  Fraction  to  its  Lowest  Terms. 

A  rational  fraction  (§  198)  is  said  to  be  in  its  loivest  terms 
when  its  numerator  and  denominator  are  prime  to  each  other 
(§  185). 

204.  Case  I.  When  the  numerator  and  denominator  can  be 
readily  factored  by  inspection. 

By  §  200,  dividing  both  terms  of  a  fraction  by  the  same  ex- 
pression, or  cancelling  common  factors  in  the  numerator  and 
denominator,  does  not  alter  the  value  of  the  fraction. 

We  then  have  the  following  rule : 

Resolve  both  numerator  and  denominator  into  their  factors, 
and  cancel  all  that  are  common  to  both. 

1.   Reduce    ^^  ^'^'^^  to  its  lowest  terms. 
40  a'b'c'd^ 

We  have,  ^^'^'^^       2^xSxa^b^cx 


40  a'b^c'd^     2^  x  5  x  a^b^c'd^ 


no       ADVANCED  COURSE  IN  ALGEBRA 

Cancelling  the  common  factor  2^  x  o?h\  we  have 
24  a^hhx       3  a'x 


40  a'b'c'd'     5  cd^ 

2. 

Reduce  J^-'^^ 

— -  to  its  lowest  terms. 
-  o 

By  §§  177  and  173, 

^-27      _(aj-3)(a^2_^3a;4-9) 

aj2  +  3a,4.9 

x^-'lx-?, 

{x-3)(x  +  l) 

x  +  1 

3. 

Reduce  ^^-^•:- 

-(^y-r  y  ^Q  ^^g  lowest  terms. 

-0? 

R^ 

7  SS  156  and  171.  - 

%x  —  bx  —  ay-\-  by  _ 

Aa-b)(x- 

-y) 

'  '  b'-a"  (b  +  a)(b-a) 

By  §  202,  the  signs  of  the  terms  of  the  factors  of  the  numerar 
tor  can  be  changed  without  altering  the  value  of  the-  fraction ; 
and  in  this  way  the  first  factor  of  the  numerator  becomes  the 
same  as  the  second  factor  of  the  denominator. 

rpi  ax  —  bx  —  ay-\-  by  _  (b  —  a)  (y  —  x)  _  y  —  x 

^^'  62_(j2        "  ~  (6  +  a)  (6  -a)~  b -j- a 

If  all  the  factors  of  the  numerator  are  cancelled,  1  remains  to  form  a 
numerator. 

If  all  the  factors  of  the  denominator  are  cancelled,  it  is  a  case  of  exact 
division. 

205.  Case  II.  When  the  numerator  and  denominator  cannot 
be  readily  factored  by  inspection. 

By  §  184,  the  H.  C.  F.  of  two  expressions  is  their  common 
factor  of  highest  degree,  having  the  numerical  coefficient  of 
greatest  absolute  value  in  its  term  of  highest  degree. 

We  then  have  the  following  rule : 

Divide  both  numerator  and  denominator  by  their  H.  C.  F. 

Ex.    Reduce  ^  ct'  -  ^^  ^'  +  7  a  -  6  ^^  .^^  ^^^^^^  ^^^^^^ 
2  a^  -  a  -  3 

By  the  rule  of  §  188,  the  H.  C.  F.  of  6  a^-ll  a^-^1  a-6  and 
2  a'  -  a  -  3  is  2  ct  -  3. 


FRACTIONS  111 

Dividing  6  a^  —  11  a^  +  7  a  —  6  by  2  a  — 3,  the  quotient  is 
3  a^  -  a  +  2. 

Dividing  2  a^  —  a  —  3  by  2  a  —  3,  the  quotient  is  a  +  1. 

6a^-lla'-{-7a-e      3a'-a-h2 


Then, 


2  a" -a- 3  a  +  1 


206.  Reduction  of  a  Fraction  to  an  Integral  or  Mixed  Expres- 
sion. 

A  Mixed  Expression  is  a  polynomial  consisting  of  a  rational 
and  integral  expression  (§  63),  together  with  one  or  more 
rational  fractions  (§  198),  each  of  which  has  letters  in  its 
denominator  when  in  its  lowest  terms  (§  203). 

Thus,  a  +-,  and  -  H — ^^ are  mixed  expressions. 

c  3       X  —  y 

1.   Eeduce i— to  a  mixed  expression. 

3  X 

By  the  Distributive  Law  for  Division  (§  100), 

6x'-\-15x-2^6x'     15a;       2  ^^^      ^ 2_ 

3a;  3a;3a;3a;  3  a;* 

A  fraction  whose  denominator  is  a  polynomial  may  be  re- 
duced to  an  integral  or  mixed  expression  by  the  operation  of 
division,  if  the  degree  (§  64)  of  the  numerator  is  not  less  than 
that  of  the  denominator. 

z.   Keduce —— — !— to  a  mixed  expression. 

4.0^  +  3)12 a^ -  8  0^  -{-  4:x -  5(3 X -  2 
12  a;^  +  9  g; 

-Sx'-Bx 
-Sx^  -  6 

—  5a;+l 

Since  the  dividend  is  equal  to  the  product  of  the  divisor  and 
quotient,  plus  the  remainder,  we  have 

12a:^-Sx'-\-4:X-5  =  (4.x'  +  3)(3  a;  -  2)  +  (-  5  a;  +  1). 


112  ADVANCED   COURSE  IN   ALGEBRA 

Then,  by  the  Distributive  Law  for  Division, 

4a^  +  3  4.0^ -{-3  Ax" +  3 

4:X^-{-3 

Then,  a  remainder  of  lower  degree  than  the  divisor  may  be 
written  over  the  divisor  in  the  form  of  a  fraction,  and  the 
result  added  to  the  quotient. 

If  the  first  term  of  the  numerator  is  negative,  as  in  Ex.  2,  it 
is  usual  to  change  the  sign  of  each  term  of  the  numerator,  chang- 
ing the  sign  before  the  fraction  (§  201). 

'  4a^  +  3  4.x^-{-3 

207.  Reduction  of  Fractions  to  their  Lowest  Common  Denomi- 
nator. 

To  reduce  fractions  to  their   Lowest   Common   Denominator 

(L.  C.  D.)  is  to  express  them  as  equivalent  fractions,  having  for 
their  common  denominator  the  L.  C.  M.  of  the  given  denomi- 
nators. 

Ex.  Reduce  — —.  — —,  and  — ^  to  their  lowest  common 
,  .     ,  3  a^6»'  2  aW'  4  a^h 

denominator. 

The  L.  C.  M.  of  3  a%\  2  ah'',  and  4  a%  is  12  a%^  (§  192). 

By  §  199,  if  the  terms  of  a  fraction  be  both  multiplied  by 

the  same  expression,  the  value  of  the  fraction  is  not  changed. 

Multiplying  both  terms  of  ^-^g  ^7  ^  ^?  ^^*^  terms  of  — ^^ 

by  6  d'h,  and  both  terms  of  j^  by  3  h\  we  have 

Uacd    18  a^hm     ^^  15  h^n 
12a^l/    12  a^b^'  12  d'l/ 

It  will  be  seen  that  the  terms  of  each  fraction  are  multi- 
plied by  an  expression,  which  is  obtained  by  dividing  the 
L.  C.  D.  by  the  denominator  of  this  fraction. 

Whence  the  following  rule. 


FRACTIONS  113 

Find  the  L.  C.  M.  of  the  given  denominators. 

Multiply  both  terms  of  each  fraction  by  the  quotient  obtained 
by  dividing  the  L.  C.  D.  by  the  denominator  of  this  fraction. 

Before  applying  the  rule,  each  fraction  should  be  reduced  to 
its  lowest  terms. 

ADDITION  AND  SUBTRACTION  OF  FRACTIONS 
208.   By§100,  ^  +  -^  =  ^  and^-^-^-^ 


a     a        a  a     a        a 

We  then  have  the  following  rules : 

To  add  two  ratioiial  fractions  which  have  a  common  denomi- 
nator, add  their  numerators^  and  make  the  residt  the  numerator 
of  a  fi'action  whose  denominator  is  the  common  denominator. 

To  subtract  two  rational  fractions  which  have  a  common  de- 
nomiiiator,  subtract  the  numerator  of  the  subtrahend  from  the 
numerator  of  the  minuend,  and  make  the  result  the  numerator  of 
a  fraction  ivhose  denominator  is  the  common  denominator. 

If  the  fractions  have  not  a  common  denominator,  it  follows 
from  §  30,  (3)  and  (4),  that  they  may  be  added,  or  subtracted, 
by  reducing  them  to  equivalent  fractions  having  their  lowest 
common  denominator  (§  207),  and  then  using  the  above  rules. 

The  final  result  should  be  reduced  to  its  lowest  terms. 

209.   1.    Simplify  A|,  +  A^. 
^     ^  4.a'b     6ab' 

The  L.  C.  D.  is  12  a^b^ 

Multiplying  the  terms  of  the  first  fraction  by  3  6^,  and  the 

terms  of  the  second  by  2  a,  we  have 

3  c        5d  ^   9b^c        Wad  ^9  b'c-^  10  ad 

4:a^b     eab^     12  a'W     12  a^l)'  12  a'b^ 

If  a  fraction  whose  numerator  is  a  polynomial  is  preceded 
by  a  —  sign,  it  is  convenient  to  enclose  the  numerator  in 
parentheses  preceded  by  a  —  sign,  as  shown  in  the  last  term 
of  the  numerator  in  equation  (A),  of  Ex.  2 ;  if  this  is  not  done, 
care  must  be  taken  to  change  the  sign  of  each  term  of  the  nu- 
merator before  combining  it  with  the  other  numerators. 


114  ADVANCED   COURSE  IN   ALGEBRA 

2.    Simplify  ^±1  +  ^^-^^^  +  ^-^^- 

Since  aj2  + 2  a;- 15  =  (:r +  5)(i»-3),  the  L.  C.  D.  is 
a^  +  2a;-15. 

Multiplying  the  terms  of  the  first  fraction  by  x  —  3,  and  the 
terms  of  the  second  by  a;  +  5,  we  have 

(x-^l)(x-3)      (x-2)(x  +  5)     2a^  +  a?-13 
x^-{-2x-15        x'-\-2x-W       a^  +  2aj-15 


(a;-|-l)(a;-3)  +  (a?-2)(a;  +  5)-(2a^  +  a;-13) 

a;2  4.2a;-15 
x^-2x-3-}-o^  +  Sx-10-2x'-x  +  13 
x'^2x-15 

^  =0(§105). 


(A) 


a;2_p2a;-15 

In  certain  cases,  the  principles  of  §§  201  and  202  enable  us 
to  change  the  form  of  a  fraction  to  one  which  is  more  con- 
venient for  the  purposes  of  addition  or  subtraction. 

3.  Simplify  _A_  + 2^. 

Changing  the  signs  of  the  terms  in  the  second  denominator, 
at  the  same  time  changing  the  sign  before  the  fraction  (§  201), 
we  have,  3     _ 26  +  a 

a-h      o?-W' 

The  L.  C.  D.  is  now  o?  -h\ 

Tiien     -? 26  +  a^3(a  +  ?>)-(2&  +  a) 

^3a  +  36-25-a^2a  +  5 

4.  Simplify  ^ : ^ ^ 

{x-y){x-z)      {y-x)(y-z)      {z-x){z-y) 

By  §  202,  we  change  the  sign  of  the  factor  y  —  x  in  the 
second  denominator,  at  the  same  time  changing  the  sign  before 
the  fraction ;  and  we  change  the  signs  of  both  factors  of  the 
third  denominator. 


FRACTIONS  115 


The  expression  then  becomes 

1      +       1 


{x-y){x-z)      (x-y){y-z)      (x-z)(y-z) 

The  L.  C.  D.  is  now  (x  —  y)(x  —  z)  {y  —  z)-,   then  the  result 

(y  —  z)  -{-  (x  —  z)  —  (x  —  y)  ^y  —  z  +  X  —  z  —  X -\- y 
(x-y)(x-z){y-z)  (x-y)(x-z)(y-z) 

2y-2z  ^  2(y-z)         .  ^  2 

(x  -y){x-  z)  (y  -z)      {x-  y)  {x  -z)(y  —  z)      {x  -y)(x-  z) 

3x-5 


5.    Simplify  2  a;  -  3 


x  +  l 


2^     3     3x-5  ^(2x-3)(x-{-l)-(3x-5) 
x-\-l  x  +  l 

^2x^-x-S-Sx  +  5^2x^-4.x-\-2 

X-^1  iC  +  l 

6.  Simplify-J-  +  -^  +  ,-^  +  ,-^- 

1  —  x     l-{-x     l  +  or     l  +  ic* 

We  first  add  the  first  two  fractions  ;  to  the  result  we  add  the 
third  fraction,  and  to  this  result  the  fourth  fraction. 

1  1      ^1+,^  +  !-^^^    2 
1-a;      l  +  x      (l-hx)(l-x)      l-x"' 

2  2     ^2(l+x')  +  2(l-a^^     4 
l-x^'^l  +  x'  {l-i-x^)(l-af)  1-x'' 

4  4     ^4(l  +  a;0+4(l-a;^)^      8 

1-a^'^l-^x'  (l-\-x')(l-x*)  1-a^' 

7.  Simplify -i- L^+     ^  ^ 


a  — 1     a  +  l      a  — 2     aH-2 

We  first  combine  the  first  two  fractions,  then  the  last  two, 
and  then  add  these  results. 

_1 1     _a  +  l-(«-l)_     2 


a-1      a  +  1      (a4-l)(ct-l)      a^-l 

1  1      ^a^2-(a-2)^     4 

a- 2      a  +  2      {a-j-2){a-2)     0^-4:' 


116  ADVANCED   COURSE  IN   ALGEBRA 


2  4      _2a2-8  +  4a2-4_     6a2-12 


4        (a2-l)(a2-4)        a^-5a2  +  4       • 

MULTIPLICATION  OF  FRACTIONS 

210.  By  §  30,  (1),  we  have  the  following  rule  for  the  product 
of  two  fractions : 

Multiply  the  numerators  together  for  the  numerator  of  the 
product,  and  the  denominators  for  its  denominator. 

211.  By  §  210,  ^  X  c  =  ^  X  J  =  y.  (1) 
Dividing  both  numerator  and  denominator  by  c  (§  200), 

fxc  =  -^.  (2) 

0  0  -T-  c 

From  (1)  and  (2),  we  have  the  following  rule  for  multiplying 
a  fraction  by  a  rational  and  integral  expression : 

If  possible,  divide  the  denominator  of  the  fraction  by  the 
expression;  otherwise,  midtiply  the  mimerator  by  the  expressioyi. 

212.  Common  factors  in  the  numerators  and  denominators 
should  be  cancelled  before  performing  the  multiplication. 

Mixed  expressions  should  be  expressed  in  a  fractional  form 
(§  209)  before  applying  the  rules. 


^^  ...   1     lOa^y  .      3M 


¥ 

10 a^y     3  6V^2  X  5  x  3  x  a^b^x'^y ^5b^x 
dbJ"  ^  4:ay~    32  X  22  X  a^ba^y'         6y  ' 
The  factors  cancelled  are  2,  3,  a^,  b,  x^,  and  y. 

2.   Multiply  together    f +  ^^      2-^^,  and  ^^. 


x^-{-x  —  6      \       x  —  Sj      x^  —  4: 

_  y4-2a;       2x-Ci-x  +  4r     x'^-9 

X ,,  X 


ic2_l_a;_6  x-3  X'-4: 

x(x-i-2)       ..a;- 2      (a;  +  3)(a;~3)  ^     x 
(x-\-S)(x-2)     ic-3^  lx  +  2){x-2)     x-2 


FRACTIONS  117 


The  factors  cancelled  are  x-\-2,  x  —  2,  x-\-S,  and  x  —  3. 
Dividing  the  denominator  by  a  —  6,  we  have 


3.   Multiply  V4^  by  «-^- 


«^  +  ^^(a-5)  =  -^  + 


4.   Multiply by  m  +  71. 

m  —  n 

Multiplying  the  numerator  by  m-\-n,  we  have 

m         .      ,     X      m^  4-  win 
X  (m  +  n) 


m—n  m—n 

DIVISION  OF  FRACTIONS 

213.  By  §  30,  (2),  we  have  the  following  rule : 

To  divide  one  fraction  by  another,  multiply  the  dividend  by  the 
divisor  inverted. 

214.  By§213,         ^^c  =  ^x-==^-  (1) 

b  b      c      be 

Dividing  both  numerator  and  denominator  by  c  (§  200), 

From  (1)  and  (2),  we  have  the  following  rule  for  dividing  a 
fraction  by  a  rational  and  integral  expression : 

If  possible,  divide  the  numerator  of  the  fraction  by  the  expres- 
sion ;  otherwise,  midtiply  the  numerator  by  the  expression. 

215     1     D"    *de    ^  ^^^   \)      ^  ^^^^  . 
5  ^y^         10  a^y 

Wehave       6  a^6  .    9  a^&^  ^  6  a^6      10  a^/ ^  4  y« 
'     5  xh/  '  10  a^/      5  0^2/'      ^  a^W       3  b^x 
Mixed  expressions  should  be  expressed  in  a  fractional  form 
(§  209)  before  applying  the  rules. 

2.   Divide    2-^^^  by  3-^^-)^. 


118  ADVANCED   COURSE  IN   ALGEBRA 


2a;-3\     [^      '^x' 


x-^1 


f) 


a;-|-2-2a;  +  3  .  3a^-3-3a^  +  13 

a^  -  1  ^  5(a;  +  1) (g;  -  1)  ^  a;  -  1 


oj  +  l         10  2x5x(a;  +  l) 


,3 


3.  Divide  '\  by  m 
Dividing  the  numerator  by  m  —  w,  we  have 

-^-— -f- (m  -  n)  ^ — ^^       ;  — 

4.  Divide  tA^  by  a  +  &. 

a-h     ^ 

Multiplying  the  denominator  by  a  +  6,  we  have 

a-h  ^  ^  a^-h^ 
If  the  numerator  and  denominator  of  the  divisor  are  exactly 
contained  in  the  numerator  and  denominator,  respectively,  of 
the  dividend,  it  follows  from  §  210  that  tlie  numerator  of  the 
quotient  may  he  ohtained  hy  dividing  the  numerator  of  the'  dividend 
by  the  numerator  of  the  divisor ;  and  the  denominator  of  the 
quotient  hy  dividing  the  denominator  of  the  dividend  by  the 
denominator  of  the  divisor. 

5.  Divide  9^-Y  by  ^l+ll. 

or  —y^  x  —  y 

We  have,       l^ziAt^Sx±2y^Sx-2y^ 
x^-y^  x-y  x-^y 

216.  By  §213,  1  =  1  x-  =  -- 

ci  a     a 

b 
Hence,  the  reciprocal  of  a  fraction  is  the  fraction  inverted. 

COMPLEX  FRACTIONS 

217.  A  Complex  Fraction  is  a  fraction  having  one  or  more 
fractions  in  either  or  both  of  its  terms. 


FRACTIONS  119 

It  is  simply  a  case  in  division  of  fractions ;  its  numerator 
being  the  dividend,  and  its  denominator  the  divisor. 

218.     1.    Simplify      ^     . 


«  X  K;r^Si  213)  = 


h  —-     ^^  ~  ^  bd  —  c^  ^      bd  —  c 

d         d 

It  is  often  advantageous  to  simplify  a  complex  fraction  by 
multiplying  its  numerator  and  denominator  by  the  L.  C.  M.  of 
their  denominators  (§  199). 

a  a 


2.   Simplify  ^^^      ^  +  ^ 


+ 


a  —  b      a  -\-b 
The  L.  C.  M.  of  a  +  &  and  a  -  6  is  (a  +  b)(a  -  b). 
Multiplying  both  terms  by  (a  -\-b)(a  —  b),  we  have 

g  (g  4-  6)  —  g  (g  —  6)  _  a^  +  g6  —  a^  +  O'b  _    2ab 
b(a-\-b)-^ala-b)~ab  +  b'-{-a^-ab~d'-{-  b^' 

1 


3.    Simplify 


1  + 


X 


in  examples  like  the  above,  it  is  best  to  begin  by  simplifying 
the  lowest  complex  fraction  ;  thus, 

1         _         1         _     x-^1     _  x-\-l 
-j    .      1      ~-.  X     ~ x  +  l  ■i-x~ 2x-^l 

X 


EXERCISE  27 

Reduce  to  their  lowest  terms  by  factoring  : 

8x3-125                                       3    a;2_ 

-9y2_z2^eyz 

2  a;3  +  x'^  -  15  X                                       '   x^ - 

-  9  2/2  +  ^2  _  2  X2; 

,         (x2-49)(x2- 16X  +  63)               ^ 

21  _  x  -  10  x2 

(x2  -  14  X  +  49)(x2  -  2  X  -  63)  15  x?/  -  20  x  -  21  ?/  +  28 


120      ADVANCED  COURSE  IN  ALGEBRA 

5    ojg  +  28  aW  +  27  b^  g        2x^-bx^-x  +  Q 

a*  +  9  ^262  +  81  ^>6  *  '   27  -  30  x  +  54  x'-^  -  8  x^* 

Reduce  to  their  lowest  terms  by  finding  the  H.  C.  F.  : 
^         10  a;2  -  7  X  -  6  g      ex^-x^-  llx  +  6 

*    4x3-4x2-5x- 1*  ■    9x8- 18x2 +  11  X -2' 

8     4  gs  4  13  a26  _  4  q&2  -  6  &3  ^^    2  x^  -  9  x^y  -  2  xy^  _  15  y^ 

8  a2  +  14  <26  _  15  62       *  '   2  x^  -  7  x'^y  -  16  xy2  +  5  y^' 

Reduce  each  of  the  following  to  a  mixed  expression  : 
jj     9x^-2x3-20  j2    m^  +  n^  ^3     12x3-3x2-22x4-8 

12  x^  m  —  n  '  3  x2  —  5 

Simplify  the  following  : 

"•   4^^x(2^  +  j,).  19.    <i±^-a  +  b. 

x2  —  2  ?/2  a  —  6 

15.    «i+64&_3  ^  ^^_    _3___A_+_^. 

a3_64&3     ^  ^  5x«     15xi/     6  2/2 


16.  ^  ~''  -;-(a  +  n).  21. 
a*  +  w* 

17.  ^^^      -^(2w-3x).  22. 
4x+5m                       ^ 

18.  3x  +  4-^^-±-^.  23 


2a  +  3      3a2+i      3a3_2 


6a 

12  a2           36 

a3 

5 
2m- 

—  + 

-  n 

2  (4  m  +  3  w) 
n2  -  4  wi2 

27x3 

+  1  , 

15  X2  -  X  - 

2 

25. 


2  X  -  3  25  x2  -  4     25  x2  -  20  X  +  4 

2  gc  -  6  gc?  -  &c  +  3  6(?  ,,  g2  -  7  g&  +  10  &2 
3gc  +  g(Z-66c-26d     10g2-3«6-&2' 
2  11 


ba ^^-^  26.     ^-^      ^+^ 


1  -  X2        1  +  X2 


„«    3m  +  1  ,   w  -  4       3 ?n2  -  2  wi  -  4 


5-2r/i  6w2-17w  +  5 

J_      J^         J 1_ 

-„      2x     3y        3x     2y  29.    -^ 1-+-^. 

^®-    4x2-9y2  +  9x2-4y2-  a^+3     x-3     x+4     x-4 

30.    (2x-l+^^^ilUfx  +  3-^^±iIV 
\  x  +  4   )      \  x  +  4   / 

«i       3g     ,     3g     ,      6  g2      ,     12  g* 


32. 


g  +  &     a  -  &     g2  +  &2     0-4  ^  ^4 

wi2  m2  2  wt^^ 


mn  +  n2     7^2  +  mn  +  w^     m*  +  w^^^  +  «* 


FRACTIONS  121 

a  —  x  a^  -r-^ 

gg   g  +  X  a^^x^  ^  \x-\ 3x  +  1 

'  a-x  a^  -x^'  '   6ic-^-17x+12      lOx^-Qx-O 

a-\-x  a^  +  x^ 

gg  h  —  c c  —  a  a  —  h 


37. 


(a-6)(a-c)      (h-c){h-a)      {c-a){c-h) 

3  3 5  n^  5  w2 

2  ri  +  1      1  -  2  n     8  w^  +  1      1  -  8  w^' 

a^      I      y L___J_ 

(X  +  1)2     (y  -  1)2     a;  4-  1     y  -  1 
1  1 


(X+l)(y-l)2        (x+l)-^(y-l) 


38        6  a2  _  a  _  2      ^^  8  a2  _  18  g  -  5  ^^^  12a2  +  7  g  -  12 
■8g2-26g  +  15       9g24-6g-8         8g2  +  6g+l* 

V.\y  -  z     xj  '  \y  +  z     x)  J  '  x'^  —  y'^  —  z^  -  2yz 
.^      x^-2x^-ix  + 


a:*-|-3a:3  -27X-81 


41     2g  -  1      2g+  1        6g-l  11 


g  -  2        g  +  2       g(2  -  g)      4  -  g2      g(g4  -  16) 


x2-(y-0)2        y2_(^_x)2  to-  /a;+l\2        /a;-l\2 

U-iJ    U  +  iy 


42.  „  ^-^  ,.-  .,  ^r"  ..•      43 


a2(l_lU&2(l_lUc2fl-l^ 
V&      c/         \c     g/        Va      &/ 


45. 


l2  -  (6  +  C)2        62  _  (c  +  a)2        C-2  -  (g  +  &)2 

1  -  X2    ,    1  +  X3 


\x-y)       \x  +  yl 


46         --      - ii^^^l—.  47     ^+^'  ^-^' 

f ^  +  y Y  I  1  I  f^  -  ?^V  1  +  x2  1  -  X8 

U-y/                   \X  +  y/  1-X2  14.341 

AQ               he             ,             ca  ,             ah 


(a  —  h){a  —  c)      (6  — c)(6  — a)      (c  — g)(c  — 6) 

49    2  +^  _L  34-x     4  +  X     2(x^-x2-19x  +  36) 
■    2-x     3-x     4-x      (x-2)(x-3)(x-4)' 


122  ADVANCED   COURSE  IN   ALGEBRA 

XI.  FRACTIONAL  AND  LITERAL  EQUATIONS 

INVOLVING  LINEAR  EQUATIONS 

219.  Suppose  that  an  equation,  with  one  unknown  number, 
X,  has  fractional  terms. 

Transpose  all  terms  to  the  first  member. 

Let  the  terms  in  the  first  member  be  then  added,  using  for  a 
common  denominator  the  L.  C.  M.  of  the  given  denominators ; 
and  let  the  resulting  fraction  be  reduced  to  its  lowest  terms. 

The  equation  will  then  be  in  the  form 

1  =  0;  ,  (1) 

where  A  and  B  are  rational  and  integral  expressions  which 
have  no  common  factor. 

If  B  contains  x,  the  given  equation  is  called  a  Fractional 
Equation. 

By  §  117,  (1)  is  equivalent  (§  114)  to  the  given  equation. 

(The  principles  demonstrated  in  §§  116  to  119,  inclusive,  and  §  122, 
hold  for  fractional  equations.) 

Equation  (1)  may  be  cleared  of  fractions  (§  121)  by  multiply- 
ing both  members  by  B,  giving  the  equation 

A  =  0.  (2) 

We  will  now  prove  equation  (2)  equivalent  to  (1). 

A 

Any  solution  of  (1),  when  substituted  for  x,  makes  —  identi- 
cally equal  to  0. 

Then,  it  must  make  A  identically  equal  to  0  (§  105). 

Then,  it  is  a  solution  of  (2). 

Again,  any  solution  of  (2),  when  substituted  for  x,  makes  A 
identically  equal  to  0. 

Now  B  cannot  equal  0  for  this  value  of  x ;  for  if  A  and  B 
became  0  for  the  same  value  of  x,  say  x  =  a,  they  would  have 
,T  —  a  as  a  common  factor  (§  141),  which  is  impossible  by  §  219. 


FRACTIONAL   AND   LITERAL   EQUATIONS         123 

Hence,  any  solution  of  (2),  when  substituted  for  a?,  makes  — 
identically  equal  to  0  (§  105). 
Then,  it  is  a  solution  of  (1). 
Therefore,  (1)  and  (2)  are  equivalent. 

220.  A  fractional  equation  may  be  cleared  of  fractions  by 
multiplying  both  members  by  any  common  multiple  of  the 
denominators ;  but  in  this  way  additional  solutions  are  intro- 
duced, and  the  resulting  equation  is  not  equivalent  to  the  first. 

Consider,  for  example,  the  equation 

1  X        _  -J 

1-f  X       l-x^~ 

Multiplying  both  members  by  1  —  x^,  the  L.  C.  M.  of  the  given 
denominators,  we  have 

l  —  x  —  x  =  l  —x"^,  ora^  —  2fl7  =  0. 

Factoring  the  first  member, 

a;(a;  -  2)  =  0 ;  and  a;  =  0  or  2  (§  182). 

If,  however,  we  multiply  both  members  of  the  given  equa- 
tion by  (1  -\-x)(l  —  X"),  we  have 

\_x'^  -  x{l-\-x)  =  (1  +  x){l  -  x"). 

Then,  1 —  x^  —  x  — x-  =  1  -\-  x  —  x- —  oi?,  ovx^  — x^  —  2x=z0. 
Factoring  the  first  member, 

x(x  +  l){x  -  2)  =  0,  and  a;  =  0,  -  1,  or  2 

This  gives  the  additional  solution  a;  =  —  1 ;  and  it  may  be 
verified  by  substitution  that  this  does  not  satisfy  the  given 
equation. 

221.  If  both  members  of  a  fractional  equation  be  multiplied 
by  any  common  multiple  of  the  denominators,  the  additional 
roots,  if  any,  introduced  must  satisfy  the  equation  formed  by 
equating  this  common  multiple  to  0. 

Thus,  in  §  220,  the  additional  solution  —  1  satisfies  the  equar 


124  ADVANCED  COURSE   IN   ALGEBRA 

If,  then,  we  reject  all  solutions  which  satisfy  the  equation 
formed  by  equating  the  common  multiple  to  0,  we  shall  retain 
the  correct  solutions. 

222.  It  follows  from  §  219  that,  if  all  the  terms  of  a  frac- 
tional equation  with  one  unknown  number  be  transposed  to 
the  first  member ;  and  all  terms  in  the  first  member  be  added, 
and  the  resulting  fraction  reduced  to  its  lowest  terms ;  and  the 
numerator  of  this  fraction  be  equated  to  0,  the  resulting  equa- 
tion is  equivalent  to  the  first. 

But  in  most  cases  the  above  is  not  the  shortest  method  of 
solution. 

By  §§  220  and  221,  we  may  multiply  the  members  of  the 
given  equation  by  the  L.  C.  M.  of  the  given  denominators  ;  and 
if  we  reject  all  solutions  which  satisfy  the  equation  formed  by 
equating  the  L.  C.  M.  to  0,  only  the  correct  solutions  will  be 
retained. 

223.  We  may  now  give  a  rule  for  solving  any  fractional  equa- 
tion, leading  to  a  linear  equation  with  one  unknown  number : 

Clear  tlie  equation  of  fractions  by  multiplying  each  term  by  the 
L.  C.  M.  of  the  given  denominators. 

Transpose  the  unknown  terms  to  the  first  member,  and  the 
known  terms  to  the  second. 

Unite  the  similar  terms,  and  divide  both  members  by  the  coeffi- 
cient of  the  unknown  number. 

2  5  2 

1.    Solve  the  equation =  — — -• 

x  —  2     £c-f2     ic^  — 4 

Multiplying  each  term  by  x^  —  4,  the  L.  0.  M.  of  the  given 
denominators,  we  have 

2(x^2)-5(x-2)  =  2. 
Or,  2x'h4.'-5x-\-10  =  2. 

Transposing,  and  uniting  terms, 

—  3a;  =  —  12,  and  ic  =  4. 

Since  4  does  not  satisfy  the  equation  a^  —  4  =  0,  it  is  the  cor- 
rect solution  (§  222). 


FRACTIONAL   AND   LITERAL  EQUATIONS         125 

If  the  denominators  are  partly  monomial  and  partly  poly- 
nomial, it  is  often  advantageous  to  clear  of  fractions  at  first 
partially ;  multiplying  each  term  of  the  equation  by  the  L.  C.  M. 
of  the  monomial  denominators. 

2.  Solve  the  equation 

6a;  +  l       2a;-4  ^2x  —  l 
15         T  0^-16  5      * 

Multiplying  each  term  by  15,  the  L.  C.  M.  of  15  and  5, 

7  X  — 16 

Transposing,  and  uniting  terms, 

^^30a;-60 
7a;-16*. 

Clearing  of  fractions,   28  x  —  64  =  30  ic  —  60. 

Then,  —  2  a?  =  4,  and  x=  —2. 

—  2  does  not  satisfy  the  equation  7  a;  —  16  =  0,  and  is  there- 
fore the  correct  root. 

If  any  fractional  terms  in  the  equation  have  the  same  de- 
nominator, they  should  be  combined  before  clearing  of  frac- 
tions. 

X  1 

3.  Solve  the  equation  — 1  =  — — -• 

x^—1  ar  —  1 

Transposing  the  term  — to  the  first  member,  and  com- 

xr  —  1 

bining  it  with  the  term  ,  we  have 

x^  —  1 

^-1  =  0,  or  ^-1=0. 
x^  —  1  x-^1 

Clearing  of  fractions,  1  —  a^  —  1  =  0,  or  x  =  0. 

0  does  not  satisfy  the  equation  oj  -|- 1  =  0,  and  is  the  correct 
root. 

If  we  solve  the  given  equation  by  multiplying  both  members  by  «*  —  1> 
we  have       x -(x^ -\)  ^\,  ov  x  -  x'^  ^  \  =  \,  ov  x  -  x'^  =  Q. 


126  ADVANCED   COURSE   IN   ALGEBRA 

Factoring  the  first  member, 

x(l  —  x)  =0  ;  and  cc  =  0  or  1. 
Now  the  value  x  =  1  satisfies  the  equation  x'^  —  1  =  0  ;  then  it  must  be 
rejected,  and  the  only  solution  is  x  =  0. 

4.    Solve  the  equation  --\ -  = — + 


x  +  S      x  +  9      a; +  10     x  i-6 
Adding  the  fractions  in  each  member,  we  have 
7  a; +  58       ^        7  a? +  58 
(a;  +  8)(a^  +  9)~(a;  +  10)(x  +  6)' 
Since  7  a;  +  58  is  a  factor  of  each  member,  we  may  place  it 
equal  to  zero  (§  182). 

Then,  7  x  +  58  =:=  0,  or  a;  =  -  — • 

The  remaining  root  is  given  by 

1  1 


(a;  +  8)  (a;  +  9)      (a;  +  10)(a;  +  6) 
or,  (x  +  8)  (aj  +  9)  =  (a?  +  10)(aj  +  6), 

or,  a^  +  17  a;  +  72  =  x^  +  1 6  a;  +  60. 

Whence,  a;  =  —  12. 

58 
Neither nor  —  12  satisfies  the  equation 

(X  +  8)  (X  +  9)  (x  +  10)  (x  +  6)  =  0. 

5.    Solve  the  equation 1 — =  2. 

2  a;  —  3      a;^  +  4 

Dividing  each  numerator  by  its  corresponding  denominator, 

we  can  write  the  equation  in  the  form 

l  +  -^  +  l-^  +  i  =  2,  or  -1 ^  =  0. 

2a;-3  a^  +  4        '        2a;-3     a;^  +  4 

Clearing  of  fractions, 

2a:2^8_2a^_5a;  +  12  =  0; 

whence,  a;  =  4. 

EXERCISE  28 

Solve  the  following  : 

J     8x-ll      7x4-4      3x-8_Q  g     12x-5      3x  +  4_4x-5 

9  12  8  ■  ■219X  +  3  7* 


FRACTIONAL  AND  LITERAL   EQUATIONS         127 


3        2x 
4a;^-( 

_2           3                        5X  +  42ic-l          1 

)         5      4x2-9                     *2x  +  3     3x-2~     6 

*•     x-b 

5x  +  4_     o                     6       ^^          7x    _12(l-a;2) 
x+8           "'                     'l-x     3  +  x     X2  4-2X-3 

7. 

2x-15x  +  6_o           23x2- 10 

3x  +  5      7-2x     "     6x2-llx-35 

8. 

3                    5                     2                     4 

a;2  _  9     x2  +  7  X  +  12     x2  -  16      x'^  -  7  x  +  12 

9. 

X2  4-3              2x-l                  1        _Q 
2x8-16      3x2  +  6x  +  12      6x-12       * 

10. 

x-3x+4_     8x+2      ^ 
x^\      x-'2.     x^-x-2 

11. 

2                1                    7               1 

2x-l      3x  +  2      6x2  +  x-2     2 

12. 

8             3            10            5 

x-3      x+7      x-9     x+2 

13. 

5                1               10               4 

2x-l      6x  +  5     3x-4     4x  +  l 

14. 

x-fl      x  +  4_x  +  2     x  +  3 
x-\      x-4      X- 2     x-3 

15. 

x+2      x-3      x+4_3 
x-3      x+4      x+2 

16. 

7  X  +  10             x2  -  3               x2  +  2      _     o 

x2  +  X  -  6      x2  -  6  X  +  8      x2  -  X  -  12  ~ 

17. 

x2-2x  +  5x2  +  3x-7_;, 
.x2-2x-3     x2  +  3x+l 

18. 

X  —  1   1  x-5     x-7  |X  —  7_^ 

x-4 

224.  Problems  involving  Fractional  Equations,  leading  to  Linear 
Equations  with  one  Unknown  Number. 

1.  A  can  do  a  piece  of  work  in  8  days,  which  B  can  perform 
in  10  days.  In  how  many  days  can  it  be  done  by  both  working 
together  ? 

Let  X  —  number  of  days  required. 

Then,  -  =  the  part  both  can  do  in  one  day. 


128      ADVANCED  COURSE  IN  ALGEBRA 

Also,  -  =  the  part  A  can  do  in  one  day, 

8 

and  —  =  the  part  B  can  do  in  one  day. 

By  the  conditions,  -  H =  -  • 

8     10     a; 

5  X  +  4  5c  =  40. 

9  a;  -  40. 

Whence,  x  —  4|,  number  Of  days  required. 

2.  The  second  digit  of  a  number  exceeds  the  first  by  2 ;  and 
if  the  number  be  divided  by  the  sum  of  its  digits,  the  quotient 
is  4^.     Find  the  number. 

Let  X  —  the  first  digit. 

Then,  x  +  2  =  the  second  digit, 

and  2  a;  +  2  =  the  sum  of  the  digits. 

The  number  itself  equals  10  times  the  first  digit,  plus  the  second  ;  then, 

10  a:  +  (x  +  2),  or  11  *  -^  2  i=  the  number. 

By  the  conditions,  li^±2  ^  34  ^ 

^  2a;  +  2        7 

77  x+ 14  =  68x4-68. 
9  X  =  54. 
Whence,  x  =  6. 

Then,  the  number  is  68. 

EXERCISE  29 

1.  The  denominator  of  a  fraction  exceeds  twice  the  numerator  by  4  ; 
and  if  the  numerator  be  increased  by  l4,  and  the  denominator  decreased 

by  9,  the  value  of  the  fraction  is  - .     Find  the  fraction. 

o 

2.  A  can  do  a  piece  of  work  in  3^  hours,  B  in  3|  hours,  and  C  in  3| 
hours.     In  how  many  hours  can  it  be  done  by  all  working  together  ? 

3.  The  second  digit  of  a  nnmber  of  two  figures  exceeds  the  first  by  5  ; 
and  if  the  number,  increased  by  1,  be  divided  by  the  sum  of  the  digits 
increased  by  2,  the  quotient  is  3.     Find  the  number. 

4.  The  numerator  of  a  fraction  exceeds  the  denominator  by  5.  If  the 
numerator  be  decreased  by  9,  and  the  denominator  increased  by  6, 
the  sum  of  the  resulting  fraction  and  the  given  fraction  is  2,  Find 
the  fraction. 


FRACTIONAL  AND  LITERAL  EQUATIONS         129 

6.  A  tank  has  three  taps.  By  the  first  it  can  be  filled  in  3  hours  10 
minutes,  by  the  second  it  can  be  filled  In  4  hours  45  minutes,  and  by  the 
third  it  can  be  emptied  in  3  hours  48  minutes.  How  many  hours  will  it 
take  to  fill  it  if  all  the  taps  are  open  ? 

6.  A  freight  train  runs  6  miles  an  hour  less  than  a  passenger  train.  It 
runs  80  miles  in  the  same  time  tliat  the  passenger  train  runs  112  miles. 
Find  the  rate  of  each  train. 

7.  The  digits  of  a  number  are  three  consecutive  numbers,  of  which  the 
middle  digit  is  the  greatest,  and  the  first  digit  the  least.    If  the  number  be 

220 
divided  by  the  sum  of  its  digits,  the  quotient  is  — '-.     Find  the  number. 

8.  A  man  walks  13J  miles,  and  returns  in  an  hour  less  time  by  a 
carriage,  whose*  rate  is  1|  times  as  great  as  his  rate  of  walking.  Find 
his  rate  of  walking. 

9.  A  vessel  runs  at  the  rate  of  llf  miles  an  hour.  It  takes  just  as 
long  to  run  23  miles  up  stream  as  47  miles  down  stream.  Find  the  rate 
of  the  stream. 

10.  A  can  do  a  piece  of  work  in  two-thirds  as  many  days  as  B,  and  B 
can  do  it  in  four-fifths  as  many  days  as  C.  Together  they  can  do  the 
work  in  3/j-  days.     In  how  many  days  can  each  alone  do  the  work  ? 

11.  The  first  digit  of  a  number  of  three  figures  is  three-fourths  the 
second,  and  exceeds  the  third  digit  by  2.  If  the  number  be  divided  by 
the  sum  of  its  digits,  the  quotient  is  38.    Find  the  number. 

12.  A  and  B  together  can  do  a  piece  of  work  in  5}  days,  B  and  C 
together  in  6f  days,  and  C  and  A  together  in  5|  days.  In  how  many  days 
can  it  be  done  by  each  working  alone  ? 

225.  Literal  Equations,  leading  to  Linear  Equations  with  one 
Unknown  Number. 

Ex.   Solve  the  equation  -^ =^+1*  =  «'  +  *'■ 

X  —  a       x-i-a       y?  —  o? 

Multiplying  each  term  by  a:^  —  a^, 

£c  (a?  -f  a)  -  (a;  -f  2  6)  (ic  -  a)  =  a^  +  5^, 

o?-\-ax  —  x^—  2  6a;  4-  aa;  +  2  a6  =  a^  +  W, 

2ax-2bx^a^-^2ab-i-b\ 
Factoring  both  members,  2  a;  (a  —  6)  =  (a  —  6)1 

Dividing  by  2  (a  -  6),  x  =  iSll^  =  ^Lul.  ■ 

^    ^     ^.         ^\  2{a-b)         2 


130  ADVANCED   COURSE  IN  ALGEBRA 


Since  ^ — —  does  not  satisfy  the  equation  x^  —  a^  =  0,  it  is  the  correct 
solution. 

EXERCISE  30 

Solve  the  following : 

1.    (x-2a-&)2-(x  +  a  +  2&)2  =  0. 
o     hx      d^  +  &2  _  g^     x{a  -  h) 

a          a^     ~  b^           b       ' 
g  5        _        2         ^ 6m 

5  X  +  2  m      4  X  —  3  TO      20  x"^  —  7  wix  —  6  w^ 
.     4  X  +  3  71     4  X  -  5  71  _  10  n^ 


X  +  2  n        Sn  —  X      x^  —  nx  —  ii  n^ 

5.  (x  +  a  -  6)3  - (X  -  a  +  6)3  =  2(a  -  6)(3 x2  +  x). 

6.  ^^-^'  +  ^^-^'^a  +  6. 

X  —  6         X  —  a 

-    X  —  g     X  —  6     X  —  c  _  ?)c(x  +  6)—  «6^  —  q'^c 

6  c  a  abc 

o        a  b         a  —  b 


x  —  a     X  —  b     X  —  c 

g  X L  =  1  _  ^^  -  2  «  -  4  a2 

'  x2  -  4  a^     2a~  x2  -  4  a2 

10      ^  —  <^        x4-a_2ax4-l^a^       ■, 
X  +  2  a     X  —  3  a     x2  —  ax  —  6  ^2 

117  3 


11. 


12. 


x-2a     6x  +  a     3x-8a     2x-3a 
4x  X  4x  X 


X— 4?i     X  +  n     X  -}-  4n     x  +  3w 
13.    (x  +  a)3+(x  +  by  -\-  (X  +  cy  =  3(x  +  a)(x  +  6)(x  +  c). 

,  ^        m  +  ?i 2  m  m  —  71     _  q 

x  +  wi  —  w     X  —  m  +  n     X  —  w  —  n 
jg         a  b  a  —  b 


X  — 26     X  — 2a     X  — a  — 6 
^g    x2  -  2  ax  +  5  a2     3  x2  +  3  ax  -  2  g^  _  ^ 

x2  -  2  ax  -  3  a2        x2  +  ax  +  2  a2 
J-    x  +  q  I  x+6  ■  X  —  g  —  6_g 

X  —  a     x  —  b      x  +  a4-6 
,Q     x+w     X  — 2w_x  +  6n     x  +  3w 


X  —  w     x  +  2?i     x  —  6  n     x  —  3  n 

19.    (x  +  2a)3+(x+ 6)3:=(2x+ 2a  4-6)3. 


FRACTIONAL   AND   LITERAL   EQUATIONS         131 

226.   Problems  involving  Literal  Equations,  leading  to  Linear 
Equations  with  one  Unknown  Number. 

Prob.     Divide  a  into  two  x^arts  such  that  m  times  the  first 
shall  exceed  n  times  the  second  by  h. 
Let  X  =  the  first  part. 

Then,  a  —  x  =  the  second  part. 

By  the  conditions,         mx  =  n{a  —  x)  +  h. 
mx  =  an  —  nx  +b. 
mx  +  nx  =  an  -\-  b. 
x{in  +  n)  =  an  -i-  b. 

Whence,  x  =  ^^1+-^,  the  first  part.  (1) 

m  +  w 

Also,  .     a-x  =  -      an  +  b_am-}-an-an-b 


m  -\-  n  m  -\-  n 

=  TILnA^  the  other  part.  (2) 

m  -i-  n 

The  results  can  be  used  as  formuloe  for  solving  any  problem  of  the 
above  form. 

Thus,  let  it  be  required  to  divide  25  into  two  parts  such  that  4  times  the 
first  shall  exceed  3  times  the  second  by  37. 

Here,  a  =  25,  m  =  4,  n  =  3,  and  b  =  37. 

Substituting  these  values  in  (1)  and  (2), 

the  first  part  ^25x3  +  37^7^+37^112^ 

7  7  7 

and  the  second  part  =  ^A^iA^IL  =  M^lIL  =  ^1  =  9. 

7  7  7 

EXERCISE  31 

1.  Divide  a  into  two  parts  whose  quotient  shall  be  m. 

2.  Two  men,  A  and  B,  a  miles  apart,  set  out  at  the  same  time,  and 
travel  towards  each  other.  A  travels  at  the  rate  of  m  miles  an  hour,  and 
B  at  the  rate  of  n  miles  an  hour.  How  far  will  each  have  travelled  when 
they  meet  ? 

3.  Divide  a  into  three  parts  such  that  the  first  shall  be  one-with  the 
second,  and  one- nth  the  third. 

4.  If  A  can  do  a  piece  of  work  in  a  hours,  B  in  6  hours,  C  in  c  hours, 
and  D  in  d  hours,  how  many  hours  will  it  take  to  do  the  work  if  all  work 
together  ? 


132  ADVANCED   COURSE  IN   ALGEBRA 

5.  What  principal  at  r  per  cent  interest  will  amount  to  a  dollars  in 
t  years  ? 

6.  In  how  many  years  will  p  dollars  amount  to  a  dollars  at  r  per  cent 
interest  ? 

7.  Divide  a  into  two  parts  such  that  one  shall  be  m  times  as  much 
above  h  as  the  other  lacks  of  c. 

8.  A  grocer  mixes  a  pounds  of  coffee  worth  m  cents  a  pound,  h  pounds 
worth  n  cents  a  pound,  and  c  pounds  worth  p  cents  a  pound.  Find  the  cost 
per  pound  of  the  mixture. 

9.  A  was  m  times  as  old  as  B  a  years  ago,  and  will  be  n  times  as  old 
as  B  in  6  years.     Find  their  ages  at  present. 

10.  If  A  and  B  can  do  a  piece  of  work  in  a  days,  B  and  C  in  6  days, 
and  A  and  C  in  c  days,  how  many  days  will  it  take  each  working  alone  ? 

227.  A  linear  equation  containing  hut  one  unknown  number 
cannot  have  more  than  one  root. 

Every  linear  equation  containing  but  one  unknown  number, 
can  be  reduced  to  the  form 

ic  =  a. 

If  possible,  let  this  equation  have  two  different  roots,  ri  and  rg. 

Then,  by  §110,  r^  =  a, 

and  ra  =  a. 

Whence,  r^  =  r^  (Ax.  4,  §  66). 

But  this  is  impossible,  since  by  hypothesis,  r^  and  r^  are  dif- 
ferent j  hence,  a  linear  equation  containing  but  one  unknown 
number  cannot  have  more  than  one  root. 


SIMULTANEOUS  LINEAR  EQUATIONS  133 

XII.    SIMULTANEOUS  LINEAR  EQUATIONS 

CONTAINING  TWO  OR  MORE  UNKNOWN  NUMBERS 

228.  An  equation  containing  two  or  more  unknown  numbers 
is  satisfied  by  an  ijidefinitely  great  number  of  sets  of  values  of 
these  numbers. 

Consider,  for  example,  the  equation  a;  +  .y  =  5. 

Putting  x  =  l,  ^Ye  have  1  -{-  y  =  5,  or  y  =  4:. 

Putting  a;  =  2,  we  have  2  +  2/  =  5,  ori/  =  3;  etc. 

Thus  the  equation  is  satisfied  by  the  sets  of  values 
a;  =  1,  2/  =  4, 
and  a;  =  2,  y  =  3  3 

and  this  could  be  extended  indefinitely,  for  we  may  give  to  x 
any  numerical  value  whatever. 

An  equation  which  has  an  indefinitely  great  number  of  solu- 
tions is  called  an  Indeterminate  Equation. 

229.  Consider  the  equations 

r       x  +  y  =  5,  (1) 

I  2a; +  22/ =  10.  (2) 

By  §  119,  equation  (2)  is  equivalent  to  (1). 
Hence,  every  solution  of  (1)  is  a  solution  of  (2),  and  every 
solution  of  (2)  is  a  solution  of  (1). 
Again,  consider  the  equations 

(x-\-y  =  5,  (3) 

U  -  2/  =  3.  (4) 

Equation  (3)  is  satisfied  by  the  set  of  values  »  =  4,  ?/  =  1 ; 
as  also  is  equation  (4). 

But  (4)  is  not  satisfied  by  every  solution  of  (3),  nor  is  (3) 
satisfied  by  every  solution  of  (4). 
Thus,  (3)  and  (4)  are  not  equivalent. 

If  two  equations,  containing  two  or  more  unknown  numbers, 
are  not  equivalent,  they  are  called  Independent 


134  ADVANCED  COURSE   IN  ALGEBRA 

230.  Consider  the  equations 

(x-^y  =  5,  (1) 

U  +  2/  =  6.  (2) 

It  is  evidently  impossible  to  find  a  set  of  values  of  x  and  y 
which  shall  satisfy  both  (1)  and  (2). 

Two  equations  which  express  incompatible  relations  between 
the  unknown  numbers  involved  are  called  Inconsistent. 

If  they  express  possible  relations  between  the  unknown 
numbers,  they  are  called  Consistent. 

231.  A  system  of  equations  is  called  Simultaneous  when  each 
contains  two  or  more  unknown  numbers,  and  every  equation 
of  the  system  is  satisfied  by  the  same  set  or  sets  of  values  of 
the  unknown  numbers. 

A  Solution  of  a  system  of  simultaneous  equations  is  a  set  of 
values  of  the  unknown  numbers  which  satisfies  every  equation 
of  the  system. 

232.  Two  systems  of  equations,  involving  two  or  more  un- 
known numbers,  are  said  to  be  equivalent  when  every  solution 
of  either  system  is  a  solution  of  the  other. 

PRINCIPLES  USED   IN  SOLVING  SIMULTANEOUS 
EQUATIONS 

233.  If  for  any  equation  of  a  system  an  equivalent  equation  be 
put,  the  resulting  system  is  equivalent  to  the  first. 

^''  |c  =  A  (1) 

be  equations  involving  two  or  more  unknown  numbers;  and 
E  =  F  an  equation  equivalent  to  (1). 
To  prove  the  system  of  equations 
jA  =  B, 

[E=F,  (2) 

equivalent  to  the  first  system. 

Since  every  solution  of  (1)  is  also  a  solution  of  (2),  every 
solution  of  the  first  system  is  a  solution  of  the  second. 


SIMULTANEOUS  LINEAR  EQUATIONS  135 

And  since  every  solution  of  (2)  is  a  solution  of  (1),  every 
solution  of  the  second  system  is  a  solution  of  the  first. 

Therefore,  the  two  systems  are  equivalent  (§  232). 

In  like  manner,  the  theorem  may  be  proved  for  a  system  of 
any  number  of  equations. 

234.  If  for  either  equation,  in  a  system  oftivo,  tve  put  an  equor 
tion  luhose  first  and  second  members  are  the  sums,  or  differences, 
respectively,  of  the  given  first  and  second  members,  the  resulting 
system  will  be  equivalent  to  the  first.   • 


Let 


\C  =  D, 


be  equations  involving  two  or  more  unknown  numbers. 
To  prove  the  system  of  equations 

equivalent  to  the  first  system. 

Any  solution  of  the  first  system,  when  substituted  for  the 
unknown  numbers,  makes  A  equal  to  B,  and  C  equal  to  D. 

It  then  makes  A+  C  equal  to  5  +  ^  (§  115,  1). 

Then  it  is  a  solution  of  the  second  system. 

Again,  any  solution  of  the  second  system  makes  A  equal  to 
B,  and  A -\-  G  equal  io  B  -\-  D. 

It  then  makes  C  equal  to  D  (§  115,  2). 

Then  it  is  a  solution  of  the  first  system. 

Therefore,  the  two  systems  are  equivalent. 

In  like  manner,  the  first  system  is  equivalent  to  the  system 
A  =  B, 
A-C  =  B-D, 

235.   It  may  be  proved,  as  in  §  234,  that  if  the  equations 

A  =  B,  C=D,E  =  F,etG.,  (1) 

involve  two  or  more  unknown  numbers,  and  any  equation  be 
replaced  by  an  equation  whose  first  member  is  the  sum  of  any 
of  the  given  first  members,  and  second  member  the  sum  of  the 
corresponding  second  members,  the  resulting  system  will  be 
equivalent  to  the  first. 


im  ADVANCED  COURSE  IN  ALGEBRA 

The  above  is  also  the  case  if  the  signs  of  both  members,  in 
any  of  the  equations  (1),  be  changed  from  +  to  — . 

For  example,  any  equation  may  be  replaced  by  the  equation 
A-\-C-E  =  B-i-D-F. 

236.  If  any  equation  of  a  system  be  solved  for  one  of  the 
unknown  numbers,  and  the  value  found  be  substituted  for  this 
unknown  number  in  each  of  the  other  equations,  the  resulting  sys- 
tem will  be  equivalent  to  the  first. 

.,  '  [A  =  B,  (1) 

^^^  \C=D,  (2) 

be  equations  involving  two  unknown  numbers,  x  and  y. 

Let  E  be  the  value  of  x  obtained  by  solving  (1) ;  and  let 
F=GhQ  the  equation  obtained  by  substituting  E  for  x  in  (2). 

To  prove  the  system  of  equations 

x  =  E,  (3) 

equivalent  to  the  first  system. 

We  know  that  (3)  is  equivalent  to  (1) ;  hence,  any  solution 
of  the  first  system  is  a  solution  of  (3)  (§  233). 

Again,  any  solution  of  the  first  system  makes  E  identically 
equal  to  x ;  and  also  makes  G  equal  to  D. 

Then  it  must  make  the  expression  obtained  by  putting  E  for 
X,  in  (7,  equal  to  the  expression  obtained  by  putting  E  for  x,  in 
D ;  that  is,  it  makes  F  equal  to  O. 

Then,  any  solution  of  the  first  system  is  also  a  solution  of 
the  second. 

Again,  since  (3)  is  equivalent  to  (1),  any  solution  of  the 
second  system  is  a  solution  of  (1). 

Also,  any  solution  of  the  second  system  makes  x  identically 
equal  to  E,  and  also  makes  F  equal  to  G. 

Then  it  must  make  the  expression  obtained  by  putting  x  for 
E,  in  F,  equal  to  the  expression  obtained  by  putting  x  for  E, 
in  G ;  that  is,  it  makes  C  equal  to  D. 

Then,  any  solution  of  the  second  system  is  also  a  solution  of 
the  first. 


SIMULTANEOUS  LINEAR  EQUATIONS  137 

Therefore,  the  two  systems  are  equivalent. 

In  like  manner,  the  theorem  may  be  proved  for  a  system  of 
any  number  of  equations,  involving  any  number  of  unknown 
numbers. 

237.  Any  equation  involving  two  unknown  numbers,  x  and 
y,  can  be  reduced  to  the  form  ax  -{-by  =  c. 

If  we  have  two  independent  simultaneous  equations  of  the 
form  ax-\-  by  =  c,  they  may  be  combined  in  such  a  way  as  to 
form  a  single  equation  involving  but  one  unknown  number. 

This^ operation  is  called  Elimination. 

There  are  three  principal  methods  of  elimination. 

I.   ELIMINATION  BY  ADDITION  OR  SUBTRACTION. 

238.  1.    Solve  the  equations 

Multiplying  (1)  by  4, 

Multiplying  (2)  by  3, 

Adding  (3)  and  (4), 

Whence, 

Substituting  a;  =  2  in  (1), 

Whence,  -  3?/  =   9,  or  y  =  -  3.  (8) 

The  above  is  an  example  of  elimination  by  addition. 

(The  principles  demonstrated  in  §§  116  to  119,  inclusive,  and  §  122, 
hold  for  equations  with  more  than  one  unknown  number.) 

By  §  119,  equation  (1)  is  equivalent  to  (3),  and  (2)  to  (4). 

Then,  by  two  applications  of  §  233,  the  given  system  is  equivalent  to 
the  system  (3)  and  (4) . 

By  §  234,  the  system  (3)  and  (4)  is  equivalent  to  the  system  (3)  and  (5). 

By  §  122,  equation  (6)  is  equivalent  to  (5)  ;  and  then  by  §  233,  the 
system  (3)  and  (5)  is  equivalent  to  the  system  (3)  and  (6),  or  to  the  sys- 
tem (1)  and  (6). 

By  §  236,  the  system  (1)  and  (6)  is  equivalent  to  the  system  (6)  and 
(7)  ;  which  by  §  233  is  equivalent  to  the  system  (6)  and  (8). 

Thus,  the  given  system  is  equivalent  to  the  system  (6)  and  (8)  ;  and 
since  no  solutions  are  introduced  nor  lost,  (6)  and  (8)  form  the  correct 
solution. 


5x-Sy  =  19. 

(1) 

,7x  +  4y=   2. 

(2) 

20a;-12y  =  76. 

(3) 

21x  +  12y=   6. 

(4) 

41  a;  =  82. 

(5) 

X:=     2. 

(6) 

10 -31/ =  19. 

(7) 

138  ADVANCED  COURSE  IN  ALGEBRA 

15x-\-Sy=        1.  (1) 


2.    Solve  the  equations 

'  10a;- 72/= -24.  (2) 

Multiplying  (1)  by  2,        S0x-{-16y=        2.  (3) 

Multiplying  (2)  by  3,        30a;-21y=  -72.  (4) 

Subtracting  (4)  from  (3),  S7  y  =  74,  and  y  =  2. 

Substituting  y  =  2m  (1),      15 a;  + 16  =  1. 

Whence,  15x=  — 15,  and  x=  —1. 

The  above  is  an  example  of  elimination  by  subtraction. 

We  speak  of  adding  a  system  of  equations  when  we  mean  placing  the 
sum  of  the  first  members  equal  to  the  sum  of  the  second  members. 

Abbreviations  of  this  kind  are  frequent  in  Algebra  ;  thus  we  speak  of 
multiplying  an  equation  when  we  mean  multiplying  each  of  its  terms. 

From  the  above  examples,  we  have  the  following  rule : 

If  necessary,  multiply  the  given  equations  by  such  numbers  as 
will  make  the  coefficients  of  07ie  of  the  unknowyi  numbers  in  the 
res  tilting  equations  of  equal  absolute  value. 

Add  or  subtract  the  resulting  equations  according  as  the  coeffi- 
cients of  equcd  absolute  vcdue  are  of  unlike  or  like  sign. 

If  the  coefficients  which  are  to  be  made  of  equal  absolute  value  are 
prime  to  each  other,  each  may  be  used  as  the  multiplier  for  the  other 
equation  ;  but  if  they  are  not  prime  to  each  other,  such  multipliers  should 
be  used  as  will  produce  their  lowest  common  multiple. 

Thus,  in  Ex.  1,  to  make  the  coefficients  of  y  of  equal  absolute  value, 
we  multiply  (1)  by  4  and  (2)  by  3;  but  in  Ex.  2,  to  make  the  coefficients 
of  X  of  equal  absolute  value,  since  the  L.C.M.  of  10  and  15  is  30,  we  mul- 
tiply (1)  by  2  and  (2)  by  3. 

II.     ELIMINATION  BY  SUBSTITUTION 

«o^     -r.       cs  ^      .^  .'        f7a;-92/=      15.  (1) 

239.   Ex.     Solve  the  equations  \ 

[82/-5a^=-17.  (2) 

Transposing  —  5  a;  in  (2),  8  ?/  =  5  a;  — 17. 

Whence,  y  =  — ~^ (3) 

8 

Substituting  in  (1),         7  a;  -  9  f^^~^'^^  =  15.  (4) 


56a;-9(5a;-17)  =  120. 

56  a; -45  03 +  153  =  120. 

11 03= -33. 

03=  -3. 

(5) 

3),,--lS-l^=-4. 

(6) 

SIMULTANEOUS   LINEAR   EQUATIONS  139 

Clearing  of  fractions, 

(5'r, 

Uniting  terms, 

Whence, 

Substituting  oj  =  —  3 

o 

By  §  236,  the  given  system  of  equations  is  equivalent  to  the  system  (3) 
and  (4)  ;  or,  since  (4)  is  equivalent  to  (5),  to  the  system  (3)  and  (5). 

By  §  236,  the  system  (3)  and  (5)  is  equivalent  to  the  system  (5)  and 
(6);  whence,  the  given  system  is  equivalent  to  the  system  (5)  and  (6). 

From  the  above  example,  we  have  the  following  rule : 

From  one  of  the  given  equations  find  the  value  of  one  of  the 

unknoivn  numbers  in  terms  of  the  other,  and  substitute  this  value 

in  place  of  that  number  in  the  other  equation. 

III.    ELIMINATION  BY  COMPARISON 


240.   Ex.    Solve  the  equations 

2o: 
.3o; 

-52/  =  -16. 

+  72/=       5. 

(1) 

(2) 

Transposing  —  5  ?/  in  (1), 

2o3  =  5?/-16. 

Whence, 

03-^2/-16. 

2 

(3) 

Transposing  7  ?/  in  (2), 

3  03=5-72/. 

Whence, 

(4) 

Equating  values  of  x,  — ^-- =  — — -^'         (5) 

Clearing  of  fractions,  15  2/  —  48  =  10  — 14  y. 

Transposing,  29  2/  =  ^^' 

Whence,  y  =  %  (6) 

Substituting  2/  =  2  in  (3),  03  =  ^-^  =  -  3.  (7) 

By  §  233,  the  given  system  of  equations  is  equivalent  to  the  system  (3) 
and  (4)  ;  or,  since,  by  Ax,  4,  §  66,  (4)  is  equivalent  to  (5),  to  the  system 

(3)  and  (5). 


140  ADVANCED   COURSE   IN   ALGEBRA 

But  (5)  is  equivalent  to  (6)  ;  so  that  the  given  system  is  equivalent  to 
the  system  (8)  and  (6). 

By  §  236,  the  system  (3)  and  (6)  is  equivalent  to  the  system  (6) 
and  (7). 

Prom  the  above  example,  we  have  the  following  rule : 

From  each  of  the  given  equations,  find  the  value  of  the  same 

unknown  number  in  terms  of  the  other,  and  place  these  values 

equal  to  each  other. 

241.  If  the  given  equations  are  not  in  the  form  ax-^by  =  c, 
they  should  first  be  reduced  to  this  form,  when  they  may  be 
solved  by  either  method  of  elimination. 

In  solving  fractional  simultaneous  equations,  we  are  liable 
to  get  results  which  do  not  satisfy  the  given  equations. 

By  §  221,  we  must  reject  any  solution  which  satisfies  the 
equation  obtained  by  equating  to  zero  the  L.  C.  M.  of  the 
given  denominators. 

■    242.   1.    Solve  the  equations 

{     7  3 


,               -^-            =0.  (1) 

[x(y-2)-y(x-5)=-13.  (2) 
Multiplying  each  term  of  (1)  by  (x  +  3)(2/  +  4), 

7y  +  28-3x-9  =  0,  ot  7y-Sx  =  -19,  (3) 
From  (2), 

xy-2x-xy-{-5y  =  -13,  or  5y-2x==-13.  (4) 

Multiplying  (3)  by  2,                         Uy-6x  =  -  38.  (5) 

Multiplying  (4)  by  3,                         Wy-6x=-39.  (6) 
Subtracting  (5)  from  (6),                                 y  =  —   1. 
Substituting  in  (4),                             —  5  -  2  a;  =  - 13. 
Whence,                                                      —2x  =  —  8,  or  a;=4. 

Since  x  =  4  and  y  =  —  1  do  not  satisfy  the  equation  (a;  +  3)  (2/  +  4)  =  0, 
the  solution  x  =  4,  y  =  —lis  correct  (§241). 

{2x-\-3y          =13.  (1) 

2.   Solve  the  equations  ]      1      ,      1     _   /^  /o^ 


SIMULTANEOUS   LINEAR   EQUATIONS 


141 


Multiplying  each  term  of  (2)  by  (x  —  2)  (y  —  3), 

y-3-\-x-2  =  0,  or  y  =  -  x  +  5.  (3) 

Substituting  in  (1),  2  x  —  3  a;  +  15  =  13,  or  a;  =  2. 

Substituting  in  (3),  2/=-2  +  5  =  3. 

Since  x  =  2  and  y  =  3  satisfy  the  equation  (x  —  2)  (y  —  3)  =  0,  the 
sohition  must  be  rejected. 

The  above  solution  satisfies  the  first  given  equation,  but  not  the  second  ; 
it  is  impossible  to  find  a  solution  which  will  satisfy  both  given  equations. 

In  solving  literal  simultaneous  linear  equations,  the  method 
of  elimination  by  addition  or  subtraction  is  usually  to  be 
preferred. 

3.   Solve  the  equations 

Multiplying  (1)  by  &', 
Multiplying  (2)  by  6, 
Subtracting, 

Whence, 

Multiplying  (1)  by  a', 
Multiplying  (2)  by  a, 
Subtracting  (3)  from  (4), 

Whence, 

^      ab'  -  a'b 

Certain  equations  in  which  the  unknown  numbers  occur  in 
the  denominators  of  fractions  may  be  readily  solved  without 
previously  clearing  of  fractions. 


'  ax  -\-by  =  c. 

(1) 

.  a'x  +  b'y  =  c'. 

(2) 

ab'x  +  bb'y  =  b'c. 

a'bx  4-  bb'y  =  be'. 

(ab'-a'b)x=b'G- 

-bc\ 

""-ab'- 

-be' 
-a'b 

aa'x  +  a'by  =  ca'. 

(3) 

aa'x  -{-  ab'y  =  c'a. 

(4) 

(ab'-a'b)y  =  c'a- 

-ca'. 

^_c'a- 

-ca' 

4.   Solve  the  equations 


Multiplying  (1)  by  5, 
Multiplying  (2)  by  3, 


10 

X 

9_ 

y 

8. 

\h 

15  _ 

y  ~ 

-1. 

50 

X 

45  _ 

y  ~ 

40. 

1- 

45 

y  ~ 

-3. 

(1) 

(2) 


142  ADVANCED   COURSE   IX    ALGEBRA 

Adding,  —  =  37. 

Whence,  74  =  37  x,  and  x  =  2. 

9 

Substituting  in  (1),  5 =  8. 

9 

Whence,  =  3,  and  yz=:  —  S. 

y 

EXERCISE   32 

Solve  the  following : 

r.08x+    .9?/ =  .048.  fx  +  y^       1  _ 

^'   I    .3x-.35?/=:.478.  5    J^~^         ^^* 

5       7  _  _  29  '   I  3x  +  8  ^  6^-1  ^ 

i-  +  ^  =  |-  r3x-4y=-ll. 

^"     '^     '  6.         2  5         , 

'  ]  5  X  +  2  ?/  ,  7  ?/  -  3  X  _  39  ■  r  5 

7  J 
rmrx  +  2/)  +  jK^-?/)  =  2.  •  I        a 

I  m- (x  +  ?/)  —  rfiix  —  y)  =  m  —  n.  [  y 

Ua  +  b)x-{-(a-h)y  =  2  (a2  +  52). 
8.  ^  fe         ^         a 

[x  —  a  —  b      y  —  a  +  h 

{     m      ,      n         .  f       3  24 


\x-a     y -\- b  ^  \2x  +  y     x-4y 


j      n      ^      m     ^^  I  __7 16_^_3^ 

[x-a     y -\- b       '  i2x  +  y     x-4:y 

2  xy  +  3  _  4y  +  5  ^ g  ^ 
a;_2         x  +  3  ^* 

'  j  2  X  +  3  y  -  1  _  3  X  -  8  ?/  ^ 25j^ 

[      2x  +  ?/  Sx-lly         (2x  +  y)(Sx-lly)' 

\  {a  -{■  b)x  -{-  {a  -  b)y  =  2  a^  -2b-\ 
12.  <      y  X  4  (tT) 


I  a  —  />      a  +  &      a^  —  b'^ 


SIMULTANEOUS   LINEAR    EQUATIONS  143 

SIMULTANEOUS    LINEAR    EQUATIONS    CONTAINING  MORE 
THAN  TWO  UNKNOWN  NUMBERS 

243.  If  we  have  three  independent  simultaneous  equations, 
containing  three  unknown  numbers,  we  may  combine  any  two 
of  them  by  one  of  the  methods  of  elimination  explained  in 
§§  238  to  240,  so  as  to  obtain  a*  single  equation  containing  only 
two  unknown  numbers. 

We  may  then  combine  the  remaining  equation  with  either 
of  the  other  two,  and  obtain  another  equation  containing  the 
same  two  unknown  numbers. 

By  solving  the  two  equations  containing  two  unknown  num- 
bers, we  may  obtain  their  values;  and  substituting  them  in 
either  of  the  given  equations,  the  value  of  the  remaining 
unknown  number  may  be  found. 

We  proceed  in  a  similar  manner  when  the  number  of  equa- 
tions and  of  unknown  numbers  is  greater  than  three. 

The  method  of  elimination  by  addition  or  subtraction  is  usu- 
ally the  most  convenient. 

If  any  equation  is  fractional,  we  should  reject  any  solution  which  satis- 
fies the  equation  obtained  by  equating  to  zero  the  L.  C.  M.  of  the  given 
denominators  (§  241). 

r    Qx-  4?/-  lz=     17.  (1) 

1.    Solve  the  equations  \    9x-  1y-lQz=     29.  (2) 

[lOx-  52/-  32=     23.  (3) 

Multiplying  (1)  by  3,       lSx-12y-21z==     51.  (4) 

Multiplying  (2)  by  2,       lSx-Uy-32z=     58.  (5) 

Subtracting,                                    2y-\-llz=-  7.  (6) 

Multiplying  (1)  by  5,       S0x-20y-S5z=     85.  (7) 

Multiplying  (3)  by  3,       3Qa;-15y-  9z=     69.  '  •  (8) 

Subtracting  (7)  from  (8),             5  y + 26  ^  =  - 16.  (9) 

Multiplying  (6)  by  5,                 10  y  +  55  z  =  -  35.  (10) 

Multiplying  (9)  by  2,  10  ?/H- 52:^^-32.  (11) 
Subtracting,  32;=— 3,  or  2  =  — 1.  (12) 
Substituting  in  (6),                       2y-ll  =  -7,  or  y=  2.       (13) 

Substituting  in  (1),                   6  .r  -  8  +  7  =  17,  or  x  =  3,        (14) 


144 


ADVANCED  COURSE  IN  ALGEBRA 


By  §§  119  and  233,  the  given  system  of  equations  is  equivalent  to  the 
system  (3),  (4),  and  (5)  ;  which,  by  §  235,  is  equivalent  to  the  system 
(3),  (4),  and  (6),  or  to  the  system  (1),  (3),  and  (6). 

But  (1)  is  equivalent  to  (7),  and  (3)  to  (8)  ;  so  that  the  given  system 
is  equivalent  to  the  system  (0),  (7),  and  (8)  ;  this,  by  §  235,  is  equiva- 
lent to  the  system  (6),  (7),  and  (9),  or  to  the  system  (1),  (6),  and  (9). 

But  the  system  (1),  (6),  and  (9)  is  equivalent  to  the  system  (1),  (10), 
and  (11)  ;  which,  by  §  235,  is  equivalent  to  the  system  (1),  (10),  and 
(12). 

The  system  (1),  (10),  and  (12)  is  equivalent  to  the  system  (1),  (G), 
and  (12)  ;  which,  by  two  applications  of  §  230,  is  equivalent  to  the  sys- 
tem (12),  (13),  and  (14). 

In  certain  cases  the  solution  may  be  abridged  by  aid  of  the 
artifice  which  is  employed  in  the  following  example. 


2.    Solve  the  equations 


2/+  z  +  u=^ 
z  -\-u-\-x  = 

Adding,  3i*  +  3aj  +  3?/  +  3  ;s  =  30. 

Whence,  u-\-x-\-y  -{-  z  =  10. 

Subtracting  (2)  from  (5),  ^^  =    3. 

Subtracting  (3)  from  (5),  x=   2. 

Subtracting  (4)  from  (5),  2/  =    1- 

Subtracting  (1)  from  (5),  2;=   4. 


6. 

(1) 

7. 

(2) 

8. 

(3) 

9. 

(4) 

(5) 


EXERCISE 

Solve  the  following : 

33 

'5x+     ?/-4;S=—  5. 

'  ax -\- by  =  {a -\-  h)c. 

1.       3x-5i/-6^=:-20. 

8. 

by  -\-  cz  =  (c  +  a)b. 

.x-Zy^^z     =-27. 

■  cz  -{-  ax  =  (6  +  c)a. 

{x-y      y  ~z      7 

3             43' 

2  a;  -  5  ?/  =  -  26. 

2. 

y-z     z  +  x_      13 
3             5             15 

4. 

7x  +  6z=  -3S. 
3             4 

z  +  X     X  —  y  _^Z 

J/-4      ^  +  2 

L     2             5     "lO' 

SIMULTANEOUS   LIKEAR  tiQUATlONS 


145 


6.    \ 


7.    ^ 


u-\-3x-2y  -  z=  -3. 
2u-x-y-^Sz  =  2S. 
u  +  x  +  Sy  -2z  z=z  -12. 
Su-2x  +  y  +  z  =  22. 

1,1,1 

-+-  +  -  =  «. 
X     y     z 

Ui  +  i  =  6. 
y     z     u 

1,1,1 

-  +  -  +  -  =  c. 

Z        U       X 

1,1,1     ^ 

-  +  -  +  -=(?. 

X       ?/       ;S 


r  a;  +  y  +  0  =  0. 
9.    j  (6  +  c)x  +  (c  +  a)y  +  («  +  6);3  =  0. 
I  6cx  +  cay  +  ahz  =  1. 


Sx~y  _  4g  -  5y  _  19 

5  '  2        ~  2* 

2a;-3g     x-iy  _7 

6  4       ~4 

4x  +  g      3y+5g_49 
3  2        ~  3  * 


r_l-+-l-=2. 

x  +  ?/      X  —  0 
6  5 


x  + 
4 


5 


=  1. 

=  2. 


0      y 


10. 


X 


y  +  b 

a+  b         c 

y       g  +  c 

6  +  c        a 
_s__x_+_a_Q 
c  +  a         6 


0. 


PROBLEMS    INVOLVING    SIMULTANEOUS    LINEAR    EQUA- 
TIONS WITH  TWO   OR  MORE  UNKNOWN  NUMBERS 

244.  In  solving  problems  where  two  or  more  letters  are 
used  to  represent  unknown  numbers,  we  must  obtain  from 
the  conditions  of  the  problem  as  many  independent  equations 
(§  229)  as  there  are  unknown  numbers  to  be  determined. 

1.  If  8  be  added  to  both  numerator  and  denominator  of 
a  fraction,  its  value  is  J;  and  if  2  be  subtracted  from  both 
numerator  and  denominator,  its  value  is  \ ;  find  the  fraction. 

Let  OJ  =  the  numerator, 

and  y  =  the  denominator. 

By  the  conditions,  ?^-±-^  =  -, 

2/4-3     3' 

a;-2^1 

2" 


and 


y-2 


Solving  these  equations,  x  =  7,  y  =  12  ;  whence,  ttie  fraction  is 


12 


146  ADVANCED   COURSE   IN    ALGEBRA 

2.  A  crew  can  row  10  miles  in  50  minutes  down  stream,  and 
12  miles  in  an  hour  and  a  half  against  the  stream.  Find  the 
rate  in  miles  per  hour  of  the  current,  and  of  the  crew  in  still 
water. 

Let  X  =  number  of  miles  an  hour  of  the  crew  in  still  water, 

and  y  =  number  of  miles  an  hour  of  the  current. 

Then,  x  +  y  =  number  of  miles  an  hour  of  the  crew  down  stream, 
and  X  —  y  =  number  of  miles  an  hour  of  the  crew  up  stream. 

The  number  of  miles  an  hour  rowed  by  the  crew  is  equal  to  the  dis- 
tance in  miles  divided  by  the  time  in  hours. 

Then,  x  +  y  =  10^^  =  12, 

6 

and  x-y  =  12-^-=   8. 

2 

Solving  these  equations,  x  =  10,  y  =  2. 

3.  The  sum  of  the  three  digits  of  a  number  is  13.  If  the 
number,  decreased  by  8,  be  divided  by  the  sum  of  its  second 
and  third  digits,  the  quotient  is  25 ;  and  if  99  be  added  to  the 
number,  the  digits  will  be  inverted.     Find  the  number. 

Let  X  =  the  first  digit, 

y  =  the  second, 
and  z  =  the  third. 

Then,  lOOx  +  lOy  +  z  =  the  number, 

and  100  z  -{-  lOy  -{■  X  =  the  number  with  its  digits  inverted. 

By  the  conditions  of  the  problem, 
x  +  y  +  z  =  lS, 

loox-f  lOy +  g-8  _gg 

and  lOOx  +  lOy  +  z  +  99  =  l00z-\-10y  +  x. 

Solving  these  equations,      x  =  2,  ?/  =  8,  0  =  3;  and  the  number  is  283. 

EXERCISE  34 

1.  If  3  be  added  to  the  numerator  of  a  certain  fraction,  and  7  sub- 
tracted  from  the  denominator,  its  value  is  -;  and  if  1  be  subtracted  from 

2 
the  numerator,  and  7  added  to  the  denominator,  its  value  is  - .     Find  the 

5 
fraction. 

2.  Find  two  numbers  such  that  one  shall  be  n  times  as  much  greater 
than  a  as  the  other  is  less  than  a  :  and  the  quotient- of  their  sum  by  their 
difference  equal  to  b. 


SIMULTANEOUS   LINEAR   EQUATIONS  147 

3.  If  the  greater  of  two  numbers  be  divided  by  the  less,  the  quotient 
is  1,  and  the  remainder  6.  And  if  the  greater,  increased  by  14,  be  divided 
by  the  less,  diminished  by  4,  the  quotient  is  5,  and  the  remainder  4.  Find 
the  numbers. 

4.  A  sum  of  money  at  simple  interest  amounted  to  $  1868.40  in  7  years, 
and  to  $2174.40  in  12  years.     Find  the  principal  and  the  rate. 

6.  A  certain  number  of  two  digits  exceeds  three  times  the  sum  of  its 
digits  by  4.  If  the  digits  be  inverted,  the  sum  of  the  resulting  number 
and  the  given  number  exceeds  three  times  the  given  number  by  2.  Find 
the  number. 

6.  A  man  invests  a  certain  sum  of  money  at  a  certain  rate  of  interest. 
If  the  principal  had  been  $1200  greater,  and  the  rate  one  per  cent  greater, 
his  income  would  have  been  increased  by  $  118.  If  the  principal  had  been 
$3200  greater,  and  the  rate  two  per  cent  greater,  his  income  would  have 
been  increased  by  $312.     What  sum  did  he  invest,  and  at  what  rate  ? 

7.  The  middle  digit  of  a  number  of  three  figures  is  one-half  the  sum  of 
the  other  two  digits.  If  the  number  be  divided  by  the  sum  of  its  digits, 
the  quotient  is  20,  and  the  remainder  9  ;  and  if  594  be  added  to  the  num- 
ber, the  digits  will  be  inverted.     Find  the  number. 

8.  A  crew  row  16|  miles  up  stream  and  18  miles  down  stream  in  9 
hours.  They  then  row  21  miles  up  stream  and  19^  miles  down  stream  in 
1 1  hours.  Find  the  rate  in  miles  an  hour  of  the  stream,  and  of  the  crew 
in  still  water. 

9.  A  and  B  can  do  a  piece  of  work  in  —  hours,  A  and  C  in  —  hours, 

21  10    ^2  ^ 

A  and  D  in  —  hours,  and  B  and  C  in  —  hours.     How  many  hours  will  it 
10  7 

take  each  alone  to  do  the  work  ? 

10.  A  and  B  run  a  race  of  280  feet.  The  first  heat,  A  gives  B  a  start 
of  70  feet,  and  neither  wins  the  race.  The  second  heat,  A  gives  B  a  start 
of  35  feet,  and  beats  him  by  6f  seconds.  How  many  feet  can  each  run  in 
a  second  ? 

11.  A,  B,  C,  and  D  play  at  cards.  After  B  has  won  one-half  of  A's 
money,  C  one-third  of  B's,  D  one-fourth  of  C's,  and  A  one-fifth  of  D's, 
tliey  have  each  $10,  except  B  who  has  $16.  How  much  had  each  at 
first  ? 

12.  The  fore-wheel  of  a  carriage  makes  a  revolutions  more  than  the 
hind-wheel  in  travelling  h  feet.  If  the  circumference  of  the  fore-wheel 
were  increased  by  one-jnth,  and  the  circumference  of  the  hind-wheel  by 
one-nth,  the  fore-wheel  would  make  c  revolutions  more  than  the  hind- 
wheel  in  travelling  d  feet.     Find  the  circumference  of  each  wheel. 


148  ADVANCED   COURSE  IN   ALGEBRA 

13.  The  sum  of  the  four  digits  of  a  number  is  14.  The  sum  of  the  last 
three  digits  exceeds  twice  the  first  by  2.  Twice  the  sum  of  the  second 
and  third  digits  exceeds  three  times  the  sum  of  the  first  and  fourth  by  3. 
And  if  2727  be  subtracted  from  the  number,  the  digits  will  be  inverted. 
Find  the  number. 

14.  A  and  B  run  a  race  from  P  to  Q  and  back ;  the  distance  from  P 
to  Q  being  108  yards.  The  first  heat,  A  reaches  Q  first,  and  meets  B  on 
his  return  at  a  point  12  yards  from  Q,  The  second  heat,  A  increases  his 
speed  by  2  yards  a  second,  and  B  by  1  yard  a  second  ;  and  now  A 
meets  B  18  yards  from  Q.    How  many  yards  can  each  run  in  a  second  ? 

15.  A  train  running  from  A  to  B,  meets  with  an  accident  which  de- 
lays it  a  hours.  It  then  proceeds  at  a  rate  one-nth  less  than  its  former 
rate,  and  arrives  at  B  &  hours  late.  Had  the  accident  occurred  c  miles 
nearer  B,  the  train  would  have  been  d  hours  late.  Find  the  rate  of  the 
train  before  the  accident,  and  the  distance  to  B  from  the  point  of 
detention. 

16.  A  man  buys  60  shares  of  stock,  part  paying  dividends  at  the  rate  of 
3|  per  cent,  and  the  remainder  at  the  rate  of  4|  per  cent.  If  the  first 
part  had  paid  dividends  at  the  rate  of  4i  per  cent,  and  the  other  at  the 
rate  of  3|  per  cent,  the  total  annual  income  would  have  been  $12  less. 
How  many  shares  of  each  kind  did  he  buy  ? 


DISCUSSION   OF   LINEAR  EQUATIONS  149 

XIII.     DISCUSSION    OF  LINEAR   EQUATIONS 

VARIABLES  AND  LIMITS 

245.  A  variable  number,  or  simply  a  variable,  is  a  number 
which  may  assume,  under  the  conditions  imposed  upon  it,  an 
indefinitely  great  number  of  different  values. 

A  constant  is  a  number  which  remains  unchanged  throughout 
the  same  discussion. 

A  limit  of  a  variable  is  a  constant  number,  the  difference 
between  which  and  the  variable  may  be  made  less  than  any 
assigned  number^  however  small,  without  ever  becoming  zero. 

In  other  words,  a  limit  of  a  variable  is  a  fixed  number  which 
the  variable  approaches  indefinitely  near,  but  never  actually 
reaches. 

246.  It  is  evident  that  the  difference  between  a  variable 
and  its  limit  is  a  variable  which  approaches  the  limit  zero. 

247.  Interpretation  of  -• 

Consider  the  series  of  fractions 
a   a   ^     a 

where  each  denominator  after  the  first  is  one-tenth  of  the  pre- 
ceding denominator. 

It  is  evident  that,  by  sufficiently  continuing  the  series,  the 
denominator  may  be  made  less  than  any  assigned  number,  how- 
ever small,  and  the  value  of  the  fraction  greater  than  any 
assigned  number,  however  great. 

In  other  words,  if  the  numerator  of  a  fraction  remains  constant, 
while  the  denominator  apinoaches  the  limit  0,  the  value  of  the 
fraction  ino^eases  ivithout  limit. 

It  is  customary  to  express  this  principle  as  follows : 
a 

5  =  '»- 


150       ADVANCED  COURSE  IN  ALGEBRA 

The  symbol  co  is  called  Infinity ;  it  simply  stands  for  that  which  is 
greater  than  any  number,  however  great. 

248.   Interpretation  of  — . 

Consider  the  series  of  fractions 
a    a      a        a 


3'  30'  300'  3000'       ' 

where  each  denominator  after  the  first  is  ten  times  the  preced- 
ing denominator. 

It  is  evident  that,  by  sufficiently  continuing  the  series,  the 
denominator  may  be  made  greater  than  any  assigned  number, 
however  great,  and  the  value  of  the  fraction  less  than  any 
assigned  number,  however  small. 

In  other  words, 

If  the  numerator  of  a  fraction  remains  constant^  while  the 
denominator  increases  without  limit ,  the  value  of  the  fraction 
ap2)7'oaches  the  limit  0. 

It  is  customary  to  express  this  principle  as  follows : 

ii  =  o. 

00 

It  must  be  clearly  understood  that  no  literal  meaning  can  be  attached 
to  such  results  as  ^ 

^=00,  and  -^  =  0; 
0         '  CO         ' 

for  there  can  be  no  such  thing  as  division,  unless  the  divisor  is  a  finite 
number. 

If  such  forms  occur  in  mathematical  investigations,  they  must  be  in- 
terpreted as  in  §§  247  and  248. 

249.   Interpretation  of  5 

By  §  44,  -  signifies  a  number  which,  when  multiplied  by  0, 
gives  0. 

But  by  §§37  and  40,  if  any  number  be  multiplied  by  0,  the 
result  is  0. 

Hence,  -r  may  be  any  number  whatever. 


DISCUSSION   OF   LINEAR  EQUATIONS  151 

For  this  reason  the  fraction  -  is  called  Indeterminate. 

250.  A  Function  of  a  number  is  any  expression  which  con- 
tains the  number. 

Thus,  the  expression  2  x^  —  3ax-\-5a^  is  a  function  of  x. 

251.  Function  Notation. 

A  function  of  x  is  often  represented  by  the  symbol /(a;)  ;  read 
"y-f unction  of  x/'  or  simply  "/-aj." 

If,  in  any  investigation, /(aj)  stands  for  a  certain  function  of 
X,  then,  whatever  value  a  may  have, /(a)  represents  the  result 
obtained  by  substituting  a  for  x  in  the  given  function. 
Thus,  if     f(x)  =x^-j-3x-2,  then 
/(3)=32  +  3. 3-2=16; 
/(_3)  =  (-3)^4-3(-3)-2=-2;  etc. 
Functions  of  x  are  also  represented  by  the  symbols  F(x),  <t>  (cc),  etc. 

THE  THEOREM  OF  LIMITS 

252.  If  two  functions  of  the  same  variable  are  so  related  that, 
as  the  variable  changes  its  value,  they  are  equal  for  every  value 
which  the  variable  can  assume,  and  each  approaches  a  certain 
limit,  then  the  two  limits  are  equal. 

Let  y  and  z  be  functions  of  a  certain  variable,  x ;  and  let  them 
be  equal  for  every  value  which  the  variable  x  can  assume,  and 
approach  the  limits  y'  and  z',  respectively. 

To  prove  that  y'  =  z'. 

Let  y^  —  y  —  m,  and  z^  —  z  =  n. 

Then,  m  and  n  are  variables  which  can  be  made  less  than 
any  assigned  number,  however  small  (§  246). 

Then,  m  —  n  is  either  zero,  or  else  a  variable  which  can  be 
made  less  than  any  assigned  number,  however  small. 

But  m  —  n  —  y^  —  y—z'-\-z  =  y^  —  z^',  for,  by  hypothesis,  y 
and  z  are  equal. 

Since  y'  —  z'  is  not  a  variable,  m  —  n  is  not  a  variable. 

Then,  m  —  n  is  0 ;  and  hence  its  equal,  y*  —  »'  is  0,  or  y'  =  «'. 


152  ADVANCED   COURSE   IN   ALGEBRA 

PROPOSITIONS  IN  REGARD  TO  LIMITS 

253.  The  limit  of  the  sum  of  a  constant  ayid  a  variable  is  the 
sum  of.  the  constant  and  the  limit  of  the  variable. 

Let  a  be  a  constant,  and  x  a  variable  whose  limit  is  x'. 

Then,  x'  —  x  can  be  made  less  than  any  assigned  number, 
however  small  (§  245). 

Whence,  (x'  +  a)  —  (oj  +  a),  which  equals  x'  —x,  can  be  made 
less  than  any  assigned  number,  however  small. 

Then,  x'  -}-  a  is  the  limit  of  x  +  a. 

254.  The  limit  of  the  sum  of  any  finite  number  of  variables  is 
the  sum  of  their  limits. 

Let  X,  y,  2,  •••,  be  variables  whose  limits  are  .t',  y\  z\  •••, 
respectively. 

Then,  x^ —x,  y'  —  y,  z' —  z,  •••,  can  be  made  less  than  any 
assigned  number,  however  small. 

Whence,  (x'  —  x) -\- (j/ —  y)  +  {z'  —  z) -\ , 

or,  (x' -\- y' -\- z' -\ )  —  (x-{-y  +  z +  '•'), 

can  be  made  less  than  any  assigned  number,  however  small. 

Then,  x'  -\- y' -\- z' -\ is  the  limit  of  x-\-j/-{-z-\ .  (1) 

255.  Any  two  corresponding  positive  signs,  in  (1),  §  254, 
may  be  changed  to  negative. 

Thus,      x'  —  y'  -^z'  -\ is  the  limit  oi  x  —  y  +  z-] . 

256.  The  limit  of  the  product  of  a  constant  and  a  variable  is 
the  constant  multiplied  by  the  limit  of  the  variable. 

Let  a,  X,  and  a?'  have  the  same  meaning  as  in  §  253. 
Then,  a{x^  —  x),  or  ax^  —  ax,  can  be  made  less  than  any  assigned 
number,  however  small. 

Whence,  ax'  is  the  limit  of  ax. 

257.  The  limit  of  the  product  of  any  number  of  variables  is  the 
product  of  their  limits. 

Let  X,  y,  z,  •••,  be  variables  having  the  limits  ic',  y',  z'^  •••, 
respectively  ;  and  let  x'  —  x  =  I,  y'  —  y  =  7n,  z'  —  z  =  n,  •••. 

Then,  /,  m,  n,  •••,  can  be  made  less  than  any  assigned  number, 
however  small. 


DISCUSSION   OF   LINEAR  EQUATIONS  153 

Now,                  x'y'z' "'  =  (x  +  ^)(y-\-7n)(z  +  7^)"• 
=  xyz h  terms  involving  I,  m,  n,  •••. 

Then,  x'y'z' ••-  —xyz---  =  terms  involving  I,  m,  n,  •••.  (1) 

The  second  member  of  (1)  can  be  made  less  than  any  assigned 
number,  however  small. 

Then,  x^y'z' "-  is  the  limit  of  xyz  •••. 

258.  Let  n  be  a  positive  integer,  and  x  a  variable  having  the 
limit  a;';  then, 

limit  of  a;"  =  limit  oi  xxxxxx  -"  to  n  factors 

=  «'  X  cc'  X  x'  X  '"  to  n  factors,  by  §  257, 

=  {xy  =  (limit  of  xy. 

259.  Tlie  limit  of  the  quotient  of  two  variables  is  the  quotient 
of  their  limits,  if  the  divisor  be  not  zero. 

Let  x  and  y  be  variables  having  the  limits  x'  and  y',  respec- 
tively ;  and  suppose  that  y'  is  not  zero. 

Let  x'  —  x  =  l,  and  y'  —  y  =  m;  then,  x  =  x'  ~l  and  y  =  y'  —  m. 

Now, 

x'     X _x'      x' —  I  _x'y' —mx' —  (x'y' —ly')  _ly' —  mx'     ,.. 

y'     y~  y'     y'  —  m~  y'{y'  —  m)  ~y'''  —  my'' 

Since  I  and  m  can  be  made  less  than  any  assigned  number, 
however  small  (§  245),  the  numerator  of  this  fraction  can  be 
made  less  than  any  assigned  number,  however  small. 

Also,  the  denominator  can  be  made  to  differ  from  y'^  by  less 
than  any  assigned  number,  however  small. 

Then,  the  fraction  (1)  can  be  made  less  than  any  assigned 
number,  however  small. 

Whence,  — ,  is  the  limit  of  -. 
2/'  y 

INTERPRETATION  OF  NEGATIVE  RESULTS 

260.  A  problem  is  said  to  be  Impossible  when  its  conditions 
cannot  be  satisfied. 

It  is  said  to  be  Indeterminate  when  the  number  of  its  solu-. 
tions  is  indefinitely  great. 


154  ADVANCED   COURSP:   IN  ALGEBRA 

261.  1.  The  length  of  a  field  is  10  rods,  and  its  breadth  8 
rods  ;  how  many  rods  must  be  added  to  the  breadth  so  that  the 
area  may  be  60  square  rods  ? 

Let  X  =  number  of  rods  to  be  added. 

By  the  conditions,  10  (8  +  x)  =  60. 

Then,  80  +  10x  =  60,  or  a:=  -  2. 

By  §  78,  adding  —  2  rods  is  the  same  thing  as  subtracting  2  rods. 

Hence,  2  rods  must  be  subtracted  from  the  breadth  in  "order  that  the 
area  may  be  60  square  rods. 

The  above  problem  is  impossible  arithmetically ;  for  the  area  of  the 
field  is  at  present  80  square  rods ;  and  it  is  impossible  to  make  it  60 
square  rods  by  adding  anything  to  the  breadth. 

If  we  should  modify  the  problem  so  as  to  read  : 

"  The  length  of  a  field  is  10  rods,  and  its  breadth  8  rods ;  how  many 
rods  must  be  subtracted  from  the  breadth  so  that  the  area  may  be  60 
square  rods  ?  " 

and  let  x  denote  the  number  of  rods  to  be  subtracted,  we  should  find  x  =  2. 
Also,  if  we  had  solved  the  given  problem  by  letting  x  denote  the  num- 
ber of  rods  to  be  subtracted,  we  should  have  found  x  =  2. 

2.  A  is  35  years  of  age,  and  B  20 ;  it  is  required  to  deter- 
mine the  epoch  at  which  A's  age  is  twice  as  great  as  B's. 

Let  us  suppose  that  the  required  epoch  is  x  years  after  the  present 
date. 

By  the  conditions,      35  +  ic  =  2  (20  +  x). 

Then,  35  +  a;  =  40  +  2  a:,  or  a;  =  -  5. 

By  §  56,  —  5  years  after  is  the  same  thing  as  5  years  before  the  present 
date. 

Therefore,  the  required  epoch  is  5  years  before  the  present  date. 

If  we  had  supposed  the  required  epoch  to  be  x  years  before  the  present 
date,  we  should  have  found  x  =  5. 

From  the  discussion  of  the  above  problems,  we  infer  that  a 
negative  result  may  be  obtained: 

1.  In  consequence  of  the  fact  that  the  problem  is  arithmeti- 
cally impossible. 

2.  In  consequence  of  a  wrong  choice  between  two  possible 
hypotheses  as  to  the  nature  of  the  unknown  number. 


DISCUSSION   OF   LINEAR  EQUATIONS  155 

In  the  first  case,  it  is  usually  possible  to  form  an  analogous 
problem,  whose  conditions  are  satisfied  by  the  absolute  value 
of  the  negative  result,  by  attributing  to  the  unknown  number 
a  quality  the  opposite  of  that  which  had  been  attributed  to  it. 

In  either  case,  a  positive  result  may  be  obtained  by  attribut- 
ing to  the  unknown  number  a  quality  the  opposite  of  that  which 
had  been  attributed  to  it;  and  the  equations  answering  to  the 
new  conditions  may  be  derived  from  the  old  equations  by 
changing  the  sign  of  the  unknown  number  wherever  it  occurs. 

Similar  considerations  hold  in  problems  involving  two  or 
more  unknown  numbers. 

A  negative  result  sometimes  indicates  that  the  problem  is 
impossible. 

3.  If  11  times  the  number  of  persons  in  a  certain  house, 
increased  by  18,  be  divided  by  4,  the  result  equals  twice  the 
number  increased  by  3 ;    find  the  number. 

Let  X  =  the  number. 

By  the  conditions,  ^^^"^^^  =  2  x  +  3. 

4 

Whence,  11  x +  18  =  8x+ 12,  and  a;  = -2. 

The  negative  result  shows  that  the  problem  is  impossible, 

262.  A  problem  may  also  be  impossible  when  the  solution 
is  fractional,  or  zero. 

1.  A  man  has  two  kinds  of  money;  dimes  and  cents.  The 
total  number  of  coins  is  23,  and  their  value  37  cents.  How 
many  has  he  of  each  ? 

Let  X  =  number  of  dimes. 

Then,  23  —  x  =  number  of  cents. 

The  X  dimes  are  worth  10  x  cents  ;  then,  by  the  conditions, 

14 
10  X  +  23  -  X  =  37  ;  and  x  =  — . 

The  fractional  result  shows  that  the  problem  is  impossible. 

2.  The  denominator  of  a  fraction  exceeds  the  numerator  by 
6 ;  and  if  2  be  added  to  the  numerator,  the  value  of  the  fraction 
is  \.     Find  the  fraction. 


156  ADVANCED   COURSE   IN   ALGEBRA 

Let  X  —  the  numerator. 

Then,"  ic  +  6  =  the  denominator. 

X  A-  2      1 
By  the  conditions,  ,  =  -  • 

Whence,  Sx  +  6=x  +  6;  and  x  =  0. 

The  result  shows  that  the  problem  is  impossible. 

THE  PROBLEM  OF  THE  COURIERS 

263.  The  discussion  of  the  following  problem  serves  to 
further  illustrate  the  interpretation  of  negative  and  zero  results, 
besides  furnishing  an  interpretation  of  infinite  and  indetermi- 
nate results. 

The  ProbTfem  of  the  Couriers.  Two  couriers,  A  and  B,  are 
travelling  along  the  same  road  in  the  same  direction,  RR',  at 
the  rates  of  m  and  n  miles  an  hour,  respectively.  .If  at  any 
time,  say  12  o'clock,  A  is  at  P,  and  B  is  a  miles  beyond  him 
at  Q,  after  how  many  hours,  and  how  many  miles  beyond  P, 
are  they  together  ? 

R  P  Q  R' 

L I I I 


Let  A  and  B  meet  x  hours  after  12  o'clock,  and  y  miles 
beyond  P. 

They  will  then  meet  y  —  a  miles  beyond  Q. 

Since  A  travels  mx  miles,  and  B  7ix  miles,  in  x  hours,  we 
l^^v®  I         y  =  mx, 

1 2/  —  (X  =  nx. 

Solving  these  equations,  we  obtain 

a  -i  am 

X  = ,  and  y  = 


m  —  n  m  —  n 

We  will  now  discuss  these  results  under  different  hypotheses. 

1.    m  >  n. 
In  this  case,  the  values  of  x  and  y  are  positive. 
This  means  that  the  couriers  meet  at  some  time  after  12,  at 
some  point  to  the  7nght  of  P. 


DISCUSSION  OF  LINEAR  EQUATIONS  157 

This  agrees  with  the  hypothesis  made ;  for  if  m  is  greater 
than  iiy  A  is  travelling  faster  than  B ;  and  he  must  overtake 
him  at  some  point  beyond  their  positions  at  12  o'clock. 

2.   m<.n. 

In  this  case,  the  values  of  x  and  y  are  negative. 

This  means  that  the  couriers  met  at  some  time  before  12,  at 
some  point  to  the  left  of  P. 

This  agrees  with  the  hypothesis  made ;  for  if  m  is  less  than 
n,  A  is  travelling  more  slowly  than  B;  and  they  must  have 
been  together  before  12  o'clock,  and  before  they  could  have 
advanced  as  far  as  P. 

3.  a  =  0,  and  m  >  n  or  m  <  n. 

In  this  case,  a;  =  0  and  y  =  0.  . 

This  means  that  the  travellers  are  together  at  12  o'clock,  at 
the  point  P. 

This  agrees  with  the  hypothesis  made ;  for  if  a  =  0,  and  m 
and  n  are  unequal,  the  couriers  are  together  at  12  o'clock,  and 
are  travelling  at  unequal  rates ;  and  they  could  not  have  been 
together  before  12,  and  will  not  be  together  afterwards. 

4.  m  =  n,  and  a  not  equal  to  0. 

In  this  case,  the  values  of  x  and  y  take  the  form  -,  and  are 
injlnite  (§  247).  ^ 

No  definite  values  can  be  assigned  to  x  and  y^  and  the  prob- 
lem is  impossible. 

This  agrees  with  the  hypothesis  made ;  for  if  m  =  n,  and  a 
is  not  zero,  the  couriers  are  a  miles  apart  at  12  o'clock,  and  are 
travelling  at  the  same  rate ;  and  they  never  could  have  been, 
and  never  will  be  together. 

Therefore,  an  infinite  result  indicates  that  the  problem  is  im- 
possible. 

In  this  case,  as  m  —  n  approaches  the  limit  0,  the  values  of  x  and  y 
increase  without  limit. 

That  is,  as  the  difference  of  the  rates  of  the  couriers  approaches  the 
limit  0,  both  the  number  of  hours  after  12  o'clock,  and  the  number  of 
miles  beyond  P,  when  A  and  B  are  together,  increase  without  limit. 


158  ADVANCED  COURSE  IN  ALGEBRA 

5.   m  =  n,  and  a  =  0. 

In  this  case,  the  values  of  cc  and  ?/  take  the  form  -,  and  are 
indeterminate  (§  249). 

This  means  that  any  value  of  x  whatever,  with  the  corre- 
sponding value  of  y,  is  a  solution  of  the  problem. 

This  agrees  with  the  hypothesis  made ;  for  if  m  =  n,  and 
a  =  0,  the  couriers  are  together  at  12  o'clock,  and  travelling  at 
the  same  rate;  and  they  always  have  been,  and  always  will 
be  together. 

Thus,  an  indeterminate  result  indicates  that  the  number  of  solu- 
tions is  indefinitely  great. 

EXERCISE  35 

Interpret  the  negative  results,  and  modify  the  enunciation  accordingly, 
in  the  following : 

1.  If  the  length  of  a  field  is  12  rods,  and  its  width  9  rods,  how  many 
rods  must  be  subtracted  from  the  width  so  that  the  area  may  be  144  square 
rods  ? 

2.  A  is  44  years  of  age,  and  B  12  years ;  how  many  years  ago  was  A 
three  times  as  old  as  B  ? 

3.  A's  assets  are  double  those  of  B.  When  A  has  gained  $250,  and  B 
$  170,  A's  assets  are  five  times  those  of  B.     Find  the  assets  of  each. 

4.  A  cistern  has  two  pipes.  When  both  are  open,  it  is  filled  in  7| 
hours ;  and  the  first  pipe  alone  can  fill  it  in  3  hours.  How  many  hours 
does  the  second  pipe  take  to  fill  it  ? 

5.  A  and  B  are  travelling  due  east  at  the  rates  of  4|  and  3^  miles  an 
hour,  respectively.  At  noon,  A  is  5  miles  due  east  of  B.  How  many 
miles  to  the  east  of  A's  position  at  noon  will  he  overtake  B  ? 

6.  A  has  $720,  and  B  .$300.  After  A  has  gained  a  certain  sum,  and 
B  has  gained  two-thirds  this  sum,  A  has  three  times  as  much  money  as  B. 
How  much  did  each  gain  ? 

In  each  of  the  following,  interpret  the  solution ; 

7.  The  number  of  apple  and  pear  trees  in  an  orchard  is  23 ;  and  seven 
times  the  number  of  apple  trees  plus  twice  the  number  of  pear  trees  equals 
82.     How  many  are  there  of  each  kind  ? 

8.  The  number  of  silver  coins  in  a  purse  exceeds  the  number  of  gold 
coins  by  3.  And  five  times  the  number  of  silver  coins  exceeds  three  times 
the  number  of  gold  coins  by  3.     How  many  are  there  of  each  kind  ? 


DISCUSSION   OF  LINEAR  EQUATIONS  159 

9.  The  numerator  of  a  fraction  is  four  times  the  denominator ;  and  if 

the  numerator  be  diminished  by  9,  and  the  denominator  by  15,  the  value 

3 
of  the  fraction  is  -  •     Find  the  fraction. 

5 

10.  A  is  a  years  old,  and  B  h  years.  After  how  many  years  will  A  be 
n  times  as  old  as  B  ? 

Discuss  the  solution  in  the  cases  when  n  =  \  and  a  is  not  equal  to  6, 
and  when  n  =  \  and  a=h. 

11.  What  number  must  be  added  to  both  terms  of  the  fraction  -  to 

c  ^ 

make  it  equal  -  ? 

Discuss  the  solution  in  the  cases  when  -  =  -  and  c  is  not  equal  to  d. 

b     d 
when  c  =  d  and  -  is  not  equal  to  -,  and  when  c  =  d  and  -  =  -. 
b  ^  d  b     d 

12.  Two  couriers,  A  and  B,  are  travelling  along  the  same  road,  in  the 
in  the  same  direction,  at  the  rates  of  m'  and  m"  miles  an  hour,  respec- 
tively. B  passes  a  certain  point  n  hours  after  A.  How  many  hours  after 
B  passes  this  point  will  he  overtake  A  ? 

Determine  for  what  values  of  the  letters  the  solution  is  positive,  nega- 
tive, zero,  impossible,  and  indeterminate,  and  discuss  the  solution  in  each 
case. 

13.  The  circumference  of  the  fore-wheel  of  a  carriage  is  a  feet,  and  of 
the  hind -wheel  b  feet.  How  far  will  the  carriage  have  travelled  when  the 
fore-wheel  has  made  n  revolutions  more  than  the  hind-wheel  ? 

Discuss  the  solution  in  the  following  cases  : 

1.  n=0,  a  and  b  unequal.     2.  a  =  b,  n  not  equal  to  zero.    3.  a  =  6,  w=0. 


INDETERMINATE    FORMS 

264.   The  indeterminate  form  -  does  not  always  represent 
a  fraction  which  may  have  any  value  whatever. 


Take,  for  example,  the  fraction . 

or —ax 

If  ic  =  a,  the  fraction  takes  the  form  -. 

Dividing  both  numerator  and  denominator  by  a;  —  a, 

x^  —  a?_x-{-a 
m?  —  ax        X     ' 
which  holds  so  long  as  x  does  not  equal  a  (§  115,  4). 


(1) 


160       ADVANCED  COURSE  IN  ALGEBRA 

The  second  member  of  (1)  approaches  the  limit  2  when  x 
approaches  the  limit  a. 

2  2 

Then,  ^  ~^    approaches  the  limit  2  when  x  approaches  the 
T    .,        x^—ax 
limit  a. 

/w2 ((2 

We  call  the  limit  approached  hy  the  fraction  — ,  when  x  ap- 

x^  —  ax 

proaches  the  limit  a,  the  value  of  the  fraction  when  x  =  a. 

In  any  similar  case,  we  divide  both  terms  of  the  fraction  by  the  ex- 
pression which  makes  each  term  vanish,  and  find  the  limit  approached  by 
the  result. 


In  the  problem  of  §  264,  the  result  -  was  obtained  in  consequence  of 

two    independent    hypotheses,   one    causing    the    numerator  to  vanish, 
and  the  other  the  denominator  ;  and  in  any  similar  case  we  should  find 

the  result  -  susceptible  of  the  same  interpretation. 

But  in  the  above  example,  the  result  -  is  obtained  in  consequence  of 
the  same  hypothesis  causing  both  numerator  and  denominator  to  vanish. 

265.   The  Indeterminate  Forms  — ,  0  x  oo,  and  go  —  oo. 

00 

1.  To  find  the  limit  approached  by  the  fraction  "^  , 
when  X  is  indefinitely  increased. 

Dividing  each  term  of  the  fraction  by  Xj  we  have 

i  +  2 

■     lim    l  +  2a;^    lim    ^ =  2±J  ('8  248^  =-• 

x^oo3  +  5x     a;  =  ao3_^g      0.+  5^  ^      5* 

X 

We  use  the  notation       ^^    for  the  words  "the  limit  when  x  is  in- 
x  =  cc  ,. 

definitely  increased  of,"  and  the  notation     **       for  the  words  "the 

X  —  a 
limit  as  x  approaches  a  of." 

In  the  above  example,  the  fraction  takes  the  indeterminate  form  — 
when  X  is  indefinitely  increased. 

In  any  similar  case,  we  divide  both  numerator  and  denominator  of  the 
fraction  by  the  highest  power  of  x. 


DISCUSSION  OF  LINEAR  EQUATIONS  161 

2.  Find  the  value  of  the  expression  (a^  +  8)[  1  H -J,  when 

The  expression  takes  the  indeterminate  form  0  x  oo  when 
x  =  -2. 
We  have 

J':,[(^  +  8)(l  +  ^)]=J-^(.'  +  8  +  »^-2x  +  4) 

^     hm     (^3^_^aj2_2a;_^12)  =  _8  +  4  +  44-12  =  12. 
X  —  —  Ji 

In  any  similar  case,  we  simplify  the  expression  as  much  as  possible 
before  finding  the  limit. 

1  2  X 

3.  Find  the  value  of  ■ when  a;  =  1. 

1  —  x     1  —  x- 

The  expression  takes  the  indeterminate  form  oo  —  oo  when 
a;  =  l. 

■j^Q^  lim   /     1  2x   \_   Urn  l-^x  —  2x 


a;  =  l\^l— a;      1  —  x'v      x  =  l      1  —  x^ 

__   lim  1  —  x  _   lim       1     _1. 

«=ll-a;2     a:  =  ll4-a;     2*  ^ 

EXERCISE  36 
Find  the  values  of  the  following  : 

1.  ^'-^^      whenx  =  4. 
a;2  _  2  ic  -  8 

2.  4  +  ox  -  3:k    ^j^gj^  ^  .g  in(jefinitely  increased. 

7  -  x  +  4  a:^ 

1  +  -^ 
3      8x^-2x-3     ^henx  =  §.  e.  -^^  when  x  =  -  1. 

12x2-25x  +  12  4  .   I       1 

X2-1 

xB  +  9x^  +  27x  +  27  ^henx=:-3.      7.  -i 12_  ^j^^^  ^  ^  2. 

x4 -18x2  +  81  x-^     x3-8 

5   4xi+2x^+2x  +  l  ^j^^^^^_l.       8  x3-3x^  +  3x-2  ^i,enx  =  2. 
8x3  +  1  2  x3-7x  +  6 

9.    (2  x2  -  6  X  -  3)  ^2  +  —^-^  when  x  =  3. 


162  ADVANCED  COURSE   IN  ALGEBRA 

DISCUSSION  OF  THE  SOLUTION  OF  A  SYSTEM  OF 
SIMULTANEOUS   LINEAR  EQUATIONS 

266.  A  system  of  simultaneous  equations  is  said  to  be  Inde- 
terminate when  the  number  of  solutions  is  indefinitely  great. 

267.  Any  system  of  two  simultaneous  linear  equations,  in- 
volving two  unknown  numbers,  can  be  reduced  to  the  form 

■  a^x-\-\y  =  c^,  (1) 

.  a^x -[- h^y  =  C2\  (2) 

where  «],  6i,  Cj,  ofg?  ^2?  ^ii^  ^s?  ii^^y  have  any  numerical  values 
whatever,  except  that  %,  hi,  a^,  and  63  cannot  be  zero. 
By  Ex.  3,  §  242,  the  solution  of  the  above  system  is 

»  =  M:::M,  and  y  =  2^h:zMi. 

aj&2  —  «2^1  tli^a  —  <^2^1 

These  fractions  have  definite  values  as  long  as  a-fi-i  does  not 
equal  a2hi. 

We  will  now  discuss  the  values  of  x  and  y  when 

ttiftg  =  Gi^i-  (A) 

(1)  If  h^Ci  —  biC2  is  not  zero,  x  is  infinite  (§  247). 

From  (A),  a,  =  ^'  (B) 

Then,  c,ai  -  c^a^  =  c^a^  -  ^^  =  f  {\c,  -  h^c^).  (C) 

Since  neither  a^,  b^,  nor  61C2  —  62^1  is  zero,  Cgai  —  Cia2  is  not 
zero. 

Whence,  y  is  also  infinite. 

By  aid  of  equation  (B),  the  given  equation  (2)  can  be  written 

^x-\-b2y  =  C2,  or  a^x -{-b^y  =  -^', 
bi  bz 

by  multiplying  each  term  by  —  • 

&2 

Thus,  the  given  equations  are  inconsistent ;  for,  by  (1),  we 
have  a^x  -^bjy  =  Cj. 


DISCUSSION  OF  LINEAR  EQUATIONS  163 

Hence,  infinite  results  show  that  the  given  equations  are  incon- 
sistent. 

(2)  If  62^1  —  &1C2  is  zero,  x  takes  the  form  - ,  and  may  have 

any  value  whatever  (§  249) ;  for  in  this  case  we  have  two  inde- 
pendent hypotheses,  one  causing  the  numerator  to  become  zero, 
and  the  other  the  denominator.     (Compare  §  264.) 

In  this  case,  by  (C),  C2a-^  —  c^az  is  also  zero ;   so  that  y  is 
indeterminate. 

From  the  equatij^ns  62^1  —  &1C2  =  0  and  C2ai  —  c^a2  =  0,  we  have 

b2  =  ^,  and  ^2  =  — • 

Then  the  given  equation  (2)  can  be  written 

^^x-\--^y  =  C2f  or  a^x -{- biy  =  Ci] 

which  is  the  same  as  (1). 

We  thus  have  a  single  equation  to  determine  two  unknown 
numbers. 

Hence,  indeterminate  results  show  that  the  given  equations  are 
not  independent. 

Similar  considerations  hold  for  any  system  of  simultaneous 
linear  equations,  involving  more  than  two  unknown  numbers. 

268.   We  will  illustrate  the  principles  of  §  267  by  an  example. 
Consider  the  system  of  equations 

f  a^x  +  h^y  +  c^z  =  (^, 
I  \  a2X  4- bzy  -\-  C2Z  =  dzf 

[  a^x  +  bsy-\-  c^z  =  d^. 
Solving,  we  find 

^  _  d^b^c^  —  t^i^gCg  +  ^^2^3^!  —  d^b-fi^  +  d^iC2  —  d^b^i ,         ,^. 
ctAcs  —  a^b^Cz  +  as^s^i  —  «2&iC3  +  «3^A  —  «3^zCi ' 
with  results  of  similar  form  for  y  and  z. 

We  will  now  use  equation  (1)  as  a  formxda  for  finding  the 
value  of  X  in  the  following  system  of  equations. 


164  ADVANCED   COURSE  IN  ALGEBRA 

i2x  +  3y-2z  =  -l, 
[     X  —  52/4-32=      6. 

Here,  ai=:8,  5i  =  —  2,  Ci  =  1,  di  ==  5,  ag  =  2,  &2  =  3,  Cg  =  —  2, 

(^2  =  —  1,  ttg  =  1,  &3  =  —  5,  C3  =  3,  dg  =  6. 

Substituting  these  values  in  (1),  we  have 

^^45 -50 +  5-6  + 24 -18^0. 
27-30-10  +  12  +  4-3     O' 

and  the  same  result  will  be  found  for  y  and  z. 

The  indeterminate  results  show  that  the  given  equations  are 
not  independent  (§  267);  this  may  be  seen  by  observing  that 
the  first  equation  is  the  sum  of  the  second  and  third. 
In  this  case,  x  may  have  any  value  whatever. 
We  will  now  apply  formula  (1)  to  the  following  system: 
2x-{-5y-Sz=     8, 
a._42/  +  22!=      3, 
3x+    y-    z  =  -2. 

Here,  ai  =  2,  &i  =  5,  Ci  =  — 3,  c?i  =  8,  a2=l,  62  =  — 4,  C2  =  2, 
(^2  =  3,  a3  =  3,  63  =  1,  C3  =  -l,  d^  =  -2. 
Substituting  these  values  in  (1),  we  have 

^^32-16-9  +  15-20  +  24^26^ 

8-4-3  +  5  +  30-36         0      "^ ' 

and  the  same  result  will  be  found  for  y  and  z. 

The  infinite  results  show  that  the  given  equations  are  incon- 
sistent (§  267) ;  this  may  be  seen  by  observing  that  the  sum 
of  the  first  two  equations  gives  3x-\-y  —  z  =  ll,  while  the  third 
requires  that  3x-{-y  —  z  should  equal  —  2. 

269.  Number  of  Solutions  of  a  System  of  Simultaneous  Linear 
Equations. 

If  we  have  a  system  of  m  independent  and  consistent  simul- 
taneous linear  equations,  involving  m  unknown  numbers,  we 
may  eliminate  m  —  1  of  the  unknown  numbers,  and  obtain  a 
single  linear  equation  involving  one  unknown  number. 


DISCUSSION  OF  LINEAR  EQUATIONS  165 

By  §  227,  the  latter  has  one  solution. 

Whence,  the  given  system-  of  equations  has  one  solution. 

And,  in  general,  a  system  of  independent  and  consistent  linear 
equations  has  a  single  solution  when  the  number  of  equations  is 
the  same  as  the  number  of  unknown  numbers. 

If  we  have  a  system  of  m  independent  linear  equations, 
involving  m  -\-n  unknown  numbers,  we  may  eliminate  m  —  1 
of  the  unknown  numbers,  and  obtain  a  single  linear  equation 
involving  the  remaining  n  + 1  unknown  numbers. 

By  §  228,  the  latter  has  an  indefinitely  great  number  of  solu- 
tions; and  hence  the  given  system  has  an  indefinitely  great 
number  of  solutions. 

And,  in  general,  a  system  of  independent  linear  equations  has 
an  indefinitely  great  number  of  solutions  when  the  number  of  equa- 
ti(Jns  is  less  than  the  number  of  unknown  numbers. 

If  we  have  a  system  oi  m-\-n  independent  linear  equations, 
involving  m  unknown  numbers,  we  may  find  a  set  of  values 
of  the  unknown  numbers  which  will  satisfy  any  m  of  the 
equations. 

But  this  set  of  values  will  not  satisfy  the  remaining  n  equa- 
tions ;  and  hence  the  given  system  has  no  solution. 

And,  in  general,  a  system  of  independent  linear  equations  has 
no  solution  when  the  number  of  equations  is  greater  than  the 
number  of  unknown  numbers. 


166 


ADVANCED  COURSE   IN  ALGEBRA 


XIV.    GRAPHICAL  REPRESENTATION 


270.  Rectangular  Co-ordinates  of  a  Point. 


1 

\(-l>,a) 

P,  (P,.a 

a 

a 

■c' 

h 

b 

•y 

^ 

N 

0           M 

'JL 

a 

a 

I 

M-bra) 

Let  XX'  and  TY'  be  straight  lines,  intersecting  at  right 
angles  at  0 ;  OY  being  above,  and  0  Y'  below,  XX ',  when  OX  is 
horizontal  and  extends  to  the  right,  and  OX'  to  the  left,  of  0. 

Let  Pj  be  any  point  in  the  plane  of  XX '  and  YY',  and  draw 
line  PjJf  perpendicular  to  XX'. 

Then,  OM  and  MP^  are  called  the  rectangular  co-ordinates  of 
Pi ;  OM  is  called  the  abscissa,  and  MPi  the  ordinate. 

The  lines  XX'  and  YY'  are  called  the  axis  of  X  and  axis  of 
Y,  respectively;  and  0  the  origin. 

We  express  the  fact  that  the  abscissa  of  a  point  is  6,  and  its 
ordinate  a,  by  saying  that,  for  the  point  in  question,  x=  b  and 
y  =  a;  or,  more  concisely,  we  speak  of  the  point  as  the  point 
(6,  a)  ;  where  the  first  term  in  parentheses  is.  understood  to  be 
the  abscissa,  and  the  second  term  the  ordinate. 

271.  Let  M  and  N  be  points  on  OX  aud  OX',  respectively, 
such  that  0M=  OJSf=  b ;  and  draw  lines  P1P4  and  P2P3  through 
M  and  JV,  respectively,  perpendicular  to  XX ',  making 

MPi=:>MP,  =  NP2  ==  i^Pa  =  a. 


GRAPHICAL   REPRESENTATION  167 

Then,  each  of  the  points  P^  P^,  Pg,  and  P4  will  have  its 
abscissa  equal  to  h,  and  its  ordinate  equal  to  a. 

To  avoid  this  ambiguity,  abscissas  measured  to  the  right  of 
O  are  considered  -|-,  and  to  the  left,  —  ;  and  ordinates  measured 
above  XX'  are  considered  -f,  and  below,—. 

Then  the  co-ordinates  of  the  points  will  be  as  follows : 
A,  (b,  a) ;  P2,  (-  6,  a)  ;  P3,  (-  6,  -  a)  ;  P„  (b,  -  a). 

It  is  understood,  in  the  above  convention  respecting  signs,  that  the 
figure  is  so  placed  that  OX  is  horizontal,  and  extends  to  the  right  of  0. 

If  a  point  lies  upon  XX,  its  ordinate  is  zero ;  and  if  it  lies 
upon  YY',  its  abscissa  is  zero. 

The  co-ordinates  of  the  origin  are  (0,  0). 

272.  Plotting  Points. 

To  plot  a  point  when  its  co-ordinates  are  given,  lay  off  the 
abscissa  to  the  right  or  left  of  0,  according  as  it  is  -f-  or  — , 
and  then  draw  a  perpendicular,  equal  in  length  to  the  ordinate, 
above  or  below  XX  according  as  the  ordinate  is  +  or  — . 

Thus,  to  plot  the  point  (—  3,  2),  lay  off  3  units  to  the  left  of 
0  upon  XX\  and  then  erect  a  perpendicular  2  units  in  length 
above  XX\ 

GRAPH  OF  A  LINEAR  EQUATION  INVOLVING  TWO 
UNKNOWN  NUMBERS 

273.  Consider  the  equation  y  =  x  +  2. 

If  we  give  any  numerical  value  to  x,  we  may,  by  aid  of  the 
relation  y  =  x-{-2,  calculate  a  corresponding  value  for  y. 


If  a;  =  0, 

2/  =  2.                 {A) 

If  a;  =  1, 

y  =  s.               (B) 

If  oj  =  2, 

2,  =  4.                  (C) 

If  a;  =  3, 

2/  =  5.                  (D) 

If  aj  =  -l. 

y  =  l.                 (E) 

•If  a;  =  -  2, 

2/  =  0.                 (F) 

If  aj  =  -3, 

2/  =  -l;  etc.     (GF) 

168 


ADVAN^CED  COURSE   IN   ALGEBRA 


Now  let  these  be  regarded  as  the  co-ordinates  of  points ;  and 
let  the  points  be  plotted,  as  explained  in  §  272. 

They  will  be  found  to  lie  on  a  certain  line,  OD,  which  is 
called  the  Graph  of  the  given  equation. 

By  assuming  fractional  values  for  a;,  we  may  obtain  intermediate  points 
of  the  graph. 

274.  We  shall  always  find  that  a  linear  equation,  involving 
two  unknown  numbers,  has  a  straight  line  for  a  graph ;  this 
may  be  proved  as  follows : 

Every  such  equation  can  be  put  in  the  form  y  =  ax-\-h. 

We  will  first  show  that  the  graph  of  2/  =  a^  is  a  straight  line. 

The  equation  y  =  ax  is  satisfied  if 
«  =  0  and  2/  =  ^  5  hence,  the  graph  of 
y  =  ax  passes  through  0. 

Let  A  and  B  be  any  other  two  points 
on  the  graph ;  draw  lines  OA  and  OB ; 
also,  lines  AC  and  BD  perpendicular  to 
OX. 

Since  A  and  B  are  on  the  graph, 

AC=axOC,  and  BD 


axOD. 


AC     BD 


since  each  of  these  equals  a. 


Then, 

'  OC     OD' 

Then,  triangles  OAC  and  OBD  are  similar ;  and  OB  coincides 
with  OA. 

Therefore,  all  points  in  the  graph  of  y  =  ax  are  in  a  straight 
line  passing  through  0. 

Now  the  graph  of  y  =  ax-\-b  can  be  obtained  from  that  of 
y  =  axhy  increasing  the  ordinate  of  each  point  of  the  graph  of 
y  =  axhj  b. 

Hence,  the  graph  of  y  —  ax-\-h  is  a  straight  line  parallel  to 
the  graph  oi  y  =  ax. 

It  follows  from  the  above  that  the  graphs  oi  y  =  ax -{- b  and  y  =  ax  +  c 
are  parallel.  , 


275.  A  straight  line  is  determined  by  any  two  of  its  points. 


GRAPHICAL   REPRESENTATION 


169 


Then,  it  is  sufficient,  when  finding  the  graph  of  a  linear  equa- 
tion involving  two  unknown  numbers,  to  find  two  of  its  points, 
and  draw  a  straight  line  through  them. 

The  points  most  easily  determined  are  those  in  which  the 
graph  intersects  the  axes. 

For  all  points  on  OX,  ?/  =  0 ;  hence,  to  find  where  the  graph 
cuts  OX,  put  y  =  0,  and  calculate  the  value  of  x. 

To  find  where  the  graph  cuts  0  Y,  put  x  =  0,  and  calculate 
the  value  of  y. 

Let  it  be  required,  for  example,  to  plot  the  graph  of 
2x  +  Sy  =  -7. 

7 
2* 


Put  2/=0;  then  2 x=—7,  and  x- 


Then  plot  A  on  OX',  -  units  to  the 
left  of  0.  ^ 


Put  a:=0;  then  3y=  —  T,  and  y= 


Then  plot  B  on  OY',  ^  units  below  O. 
o 

Draw  the  straight  line  AB ;  this  is  the  required  graph. 

The  above  method  cannot,  of  course,  be  used  for  a  straight  line  passing 
through  the  origin,  nor  for  the  equations  of  §  276. 

276.   Consider  the  equation  y  =  5. 

This  means  that  every  point  in  the 
graph  has  its  ordinate  equal  to  5. 

Then  the  graph  is  the  straight  line  AB, 
parallel  to  XX',  and  5  units  above  it. 

In  like  manner,  the  graph  of  x  =  —  3 
is  the  straight  line  CD,  parallel  to  YY', 
and  3  units  to  the  left  of  it. 

The  graph  oiy  =  Ois  the  axis  of  X,  and  the  graph  of  x  =  0  is  the  axis 
of  Y. 

EXERCISE  37 


O 

Y 

JL 

y' 

0 

D 

r 

Plot  the  graphs  of  the  following  : 

1.  3x  +  2y  =  6.       3.   a;  =  2.  5.   2  a;  =  3  y. 

2.  aj  -  4  2/  =  4.         4.   y  =  -  4.      6.   a;  +  y  =  0. 


7.  16a;-27y  =  -72. 

8.  8  a;  +  16  y  =  -  6. 


170 


ADVANCED   COURSE   IN   ALGEBRA 


INTERSECTIONS    OF    GRAPHS 

277.   Consider  the  equations 
3  a;  -  ?/  =  -  9.     (AB) 
x-{-2y=      4.     (CD) 

Let  AB  be  the  graph  of  3  a;  —  ?/  =  —  9, 
and  CD  the  graph  of  x-\-2y  =  4:. 

Let  AB  and  CD  intersect  at  E. 

Since  E  lies  on   each   graph,  its  co- 
ordinates must  satisfy  both  given  equa- 
tions ;  hence,  to  find  the  co-ordinates  of  E,  we  solve  the  given 
equations. 

In  this  case,  the  solution  is  x  =  —  2,  y  =  3;  and  it  may  be 
verified  in  the  figure  that  these  are  the  co-ordinates  of  E. 

Hence,  if  the  graphs  of  any  tivo  linear  equations,  ivith  two 
unknown  numbers,  intersect,  the  co-ordinates  of  the  point  of  inter- 
section form  a  solution  of  the  system  of  equations  represented  by 
the  graphs. 

EXERCISE  38 


Verify  the  principle  of  §  277  in  the  following  systems  : 

r  5  X  -  4  y  =  0. 
y  =  -o.  *l7x  +  6y=:-29. 

3x4-7^  =  5. 
8  a;  +  3  2/  =  -  18. 


r  4  X  +  5  y  =  24. 
\sx-2y=-6. 


4. 


r  9  X  +  14  y  =  -  25. 
1  3  X  -  4  V  =  22. 


'    278.   Graphs  of  Inconsistent  and  Indeterminate  Linear  Equa- 
tions, with  two  Unknown  Numbers. 

Consider  the  equations 

(3x-2y=      5.     (AB) 
[6x-4:y=-7.     (CD) 

The  equations  can  be  written, 

3        5        -,         3,7 

V  =  -  X ,  and  y  =  -x-{--  ; 

2        2  2        4 

and  it  was  shown  in  §  274  that  the  graphs 
of  two  equations  of  this  form  are  parallel. 


GRAPHICAL   REPRESENTATION 


171 


The  given  equations  are  inconsistent ;  and  we  shall  always 
find  that  two  inconsistent  equations,  with  two  unknown  num- 
bers, are  represented  by  parallel  graphs. 

Again,  consider  the  equations 
3x-2y=:   5. 
\6x  —  4:y  =  10. 

In  this  case,  the  graphs  coincide. 

The  given  equations  are  not  independent; 
and  in  any  similar  case,  we  should  find 
that  the  graphs  were  coincident. 

279.  Graphical  Representation  of  Linear  Expressions  involving 
one  Unknown  Number. 

Consider  the  expression  3  x  +  5. 

Put  2/  =  3  a;  4-  5  ;  and  let  the  graph  of 
this  equation  be  found  as  in  §  275. 


r 

:  / 

1 

B 

i 

0    ■ 

Y' 

Putting  2/  =  0,  x= ;  then  the  graph 

cuts  XX'  ~  units  to  the  left  of  0. 
3 

Putting  a;  =  0,   ?/  =  ^  5    then  the   graph 
cuts  TY'  5  units  above  0. 

The  graph  is  the  straight  line  AB. 

It  was  shown  in  §  274  that  the  graph  of  a  linear  equation,  with  two 
unknown  numbers,  is  a  straight  line. 

280.   Graphical  Representation  of  Roots  (§  110). 
Consider  the  equation  ax  -\-  h  —  Q.  (1) 

To  find  the  graph  of  the  first  member,  put  y  =  ax-\-h.       (2) 
The  abscissa  of  the  point  in  which  this  graph  intersects  OX, 

must,  when  substituted  for  x  in  (2),  make  2/  =  0. 

Then,  it  makes  the  first  member  of  (1)  equal  to  zero,  and  is 

therefore  a  root  of  the  given  equation. 

Hence,  the  abscissa  of  the  point  in  which  the  graph  of  the  first 
member  of  any  linear  equation,  with  one  unknown  number,  inter- 
sects XX',  is  the  root  of  the  equation. 


172  ADVANCED   COURSE   IN  ALGEBRA 

Consider,  for  example,  the  equation  3  x  +  5  =  0. 
The  graph  of  the  first  member  was  found  in  §  279. 
This  intersects  XX'  at  the  point  -4,  whose  abscissa  is  — ;  and  the 
root  of  the  equation  is  —  — 

EXERCISE  39 

Verify  the  principles  of  §  278  in  the  four  following  systems  : 


r3x  +  4 
l3x  +  4 


2/=      16.  (2x-   7y  =  U. 

y=  -16.  ■     l4x-14?/  =  28. 

(2x-    6y=   0.  (   5x+   6y=15. 

lex -15?/ =  30.  ■    ll5x+18?/  =  45. 

Plot  the  graphs  of  the  first  members  of  the  following  equations,  and 
in  each  case  verify  the  principle  of  §  280 : 

6.   2x  +  7  =  0.  6.   5x-4  =  0. 


INVOLUTION  AND  EVOLUTION  173 

XV.    INVOLUTION  AND  EVOLUTION 

INVOLUTION 

281.  Involution  is  the  process  of  raising  an  expression  to 
any  power  whose  exponent  is  a  positive  integer. 

We  have  already  given  (§  130)  a  rule  for  raising  a  rational 
and  integral  monomial  to  any  power  whose  exponent  is  a 
positive  integer. 

282.  Any  Power  of  a  Fraction. 

We  have         f  -  Y*  =  ?  x  ?  X  •  •  •  to  n  factors 


bj       b      b 

_axax  • "  to  n  factors _ ct" 
b  X  b  X  •"  to  n  factors      6" 

Then,  a  fraction  may  be  raised  to  any  power  whose  exponent  is 
a  positive  integer  by  raising  both  numerator  and  denominator  to 
the  required  poioer,  and  dividing  the  first  result  by  the  second. 

THE  BINOMIAL  THEOREM 

283.  A  Series  is  a  succession  of  terms. 

A  Finite  Series  is  one  having  a  limited  number  of  terms. 
An  Infinite  Series  is  one  having  an  unlimited  number  of  terms. 

284.  In  §§  131  and  135,  we  gave  rules  for  finding  the  square 
or  cube  of  any  binomial. 

The  Binomial  Theorem  is  a  formula  by  means  of  which  any 
power  of  a  binomial  may  be  expanded  into  a  series. 

In  the  present  chapter,  we  shall  consider  those  cases  only  in 
which  the  exponent  is  a  positive  integer. 

285.  Proof  of  the  Theorem  for  a  Positive  Integral  Exponent. 
The  following  are  obtained  by  actual  multiplication. 


174  ADVANCED   COURSE   IN   ALGEBRA 

(a  -\- xy  =  a^ -{- 2  ax -\- x^ ; 

(a-{-xY  =  a^-\-3a'x-\-3ax'  +  x^; 

(a  -{- xy  =  a^ -\- 4:  a?x  -)-  6  a-x^  +  4  aa^  +  ic^ ;  etc. 

In  these  results,  we  observe  the  following  laws : 

1.  The  number  of  terms  is  greater  by  1  than  the  exponent  of 
the  binomial. 

2.  The  exponent  of  a  in  the  first  term  is  the  same  as  the 
exponent  of  the  binomial,  and  decreases  by  1  in  each  succeed- 
ing term. 

3.  The  exponent  of  x  in  the  second  term  is  1,  and  increases 
by  1  in  each  succeeding  term. 

4.  The  coefficient  of  the  first  term  is  1,  and  the  coefficient  of 
the  second  term  is  the  exponent  of  the  binomial. 

5.  If  the  coefficient  of  any  term  be  multiplied  by  the  exponent 
of  a  in  that  term,  and  the  result  divided  by  the  exponent  of  x 
in  the  term  increased  by  1,  the  quotient  will  be  the  coefficient 
of  the  next  following  term. 

We  will  now  prove  by  Mathematical  Induction  (Note,  §  134) 
that  these  laws  hold  for  any  positive  integral  power  of  a-\-x. 

Assume  the  laws  to  hold  for  (a  +  x)",  where  n  is  any  positive 
integer. 

Then,     (a  +  xy  =  a"  +  wa^-^o;  +  ^^^^  ~  ^^  a^-^x^  +  •  •  ••  (1) 

Let  P,  Q,  and  R  denote  the  coefficients  of  the  terms  involv- 
ing a"~'"a;'",  a""''~V+^,  and  a^'^^'V^^,  respectively,  in  the  second 
member  of  (1)  ;  thus, 

{a -\- xy  =  or- -\- noT-'^x -\-  ♦•• 

-f-  Par-'^x^  -t-  Qa"-'-V+^  -f  Ita''-'-^x'+^  +  ....      (2) 

Multiplying  both  members  by  a  +  cc,  we  have 

(a  -f  x)  "+^  =  a"+^  -f  na^x  -\ \-  Qa"-''aj''+i  +  -Ra"-''-^x'+2  ^ 

-f  a"a;  H h  Pof"~'x!'+^  +  Qa"-'-V+2  -j 

=  «"+'  + (ji  +  1)  a'^a;  H 

+  (P  +  Q)a^-''x'+'  +(Q  +  i?)a"-'- V+2 -f  ....  (3) 


INVOLUTION   AND  EVOLUTION  175 

This  result  is  in  accordance  with  the  second^  third,  and  fourth 
laws. 

Since  the  fifth  law  is  assumed  to  hold  with  respect  to  the 
second  member  of  (2),  we  shall  have 

r +  1  r  +  2 

Therefore, 

Q(n-r-l)      Q(n  +  1) 
Q-\-B_  r  +  2        _     r  +  2     _n-7 

P+Q~      Q(r  +  1)  ^  ~Q(n  +  l)~r  +  2' 

n  —  r  n  —  r 

n  —  r 


Whence,  Q  +  i2  =  (P  +  Q) 


r  +  2 


But  n  —  r  is  the  exponent  of  a  in  that  term  of  (3)  whose  coeffi- 
cient is  P-\-Qi  and  r  +  2  is  the  exponent  of  x  increased  by  1. 

Therefore,  \hQ  fifth  law  holds  with  respect  to  (3). 

Hence,  if  the  laws  hold  for  any  power  of  a  +  a?  whose  expo- 
nent is  a  positive  integer,  they  also  hold  for  a  power  whose 
exponent  is  greater  by  1. 

But  the  laws  have  been  shown  to  hold  for  (a  +  xy,  and  hence 
they  hold  for  (a  +  xf ;  and  since  they  hold  for  (a  +  xy,  they 
hold  for  (a  +  xy ;  and  so  on. 

Hence,  they  hold  when  the  exponent  is  any  positive  integer. 

By  aid  of  the  fifth  law,  the  coefficients  of  the  successive 
terms  after  the  second,  in  the  second  member  of  (2),  may  be 
readily  found ;  thus, 

(a  +  xf  =  a**  +  na^'-'^x  +  ^^!^  ""-*•)  a""  V 

1  •  i& 

+  "^"7^iV^^«'-V+-.  (4) 

This  result  is  called  the  Binomial  TJieorem. 

In  place  of  the  denominators  1  •  2,  1  •  2  •  3,  etc.,  it  is  customary  to 
write  [2,  [3,  etc.  The  symbol  [w,  read  '^factorial  n,"  signifies  the  product 
of  the  natural  numbers  from  1  to  n  inclusive. 


176  ADVANCED   COURSE   IN   ALGEBRA 

286.  In  expanding  expressions  by  the  Binomial  Theorem,  it 
is  convenient  to  obtain  the  exponents  and  coefficients  of  the 
terms  by  aid  of  the  laws  of  §  285,  which  have  been  proved  to 
hold  for  any  positive  integral  exponent. 

1.  Expand  {a  +  xy. 

The  exponent  of  a  in  the  first  term  is  5,  and  decreases  by  1 
in  each  succeeding  term. 

The  exponent  of  x  in  the  second  term  is  1,  and  increases  by 
1  in  each  succeeding  term. 

The  coefficient  of  the  first  term  is  1 ;  of  the  second,  5. 

Multiplying  5,  the  coefficient  of  the  second  term,  by  4,  the 
exponent  of  a  in  that  term,  and  dividing  the  result  by  the 
exponent  of  x  increased  by  1,  or  2,  we  have  10  as  the  coefficient 
of  the  third  term ;  and  so  on. 

Then,  (a  +  ic)^  =  a^  +  5  a*x  +  10  aV  + 10  aV  +  5  ax^-[-  x^. 

It  will  be  observed  that  the  coefficients  of  terms  equally  distant  from 
the  ends  of  the  expansion  are  equal ;  this  law  will  be  proved  in  §  288. 

Thus  the  coefficients  of  the  latter  half  of  an  expansion  may  be  written 
out  from  the  first  half. 

If  the  second  term  of  the  binomial  is  negative,  it  should  be 
enclosed,  negative  sign  and  all,  in  parentheses  before  applying 
the  laws ;  in  reducing,  care  must  be  taken  to  apply  the  princi- 
ples of  §  130. 

2.  Expand  (1  -  xf. 

(l-Xy=:\:i  +  (-X)f 

=  16  +  6  .  1^ .  (-  oj)  +  15  .  1^ .  (-  xy  +  20  .  13 .  (-  xy 

-h^5  'V '  (-xy  +  6  '1 '  (-xy-\-{-xy 

=  l-6x-\-15a^-20a^A-15x^-6a^-^x^ 

If  the  first  term  of  the  binomial  is  a  number  expressed  in  Arabic 
numerals,  it  is  convenient  to  write  the  exponents  at  first  without  reduc- 
tion ;  the  result  should  afterwards  be  reduced  to  its  simplest  form. 

If  either  term  of  the  binomial  has  a  coefficient  or  exponent 
other  than  unity,  it  should  be  enclosed  in  parentheses  before 
applying  the  laws. 


INVOLUTION  AND  EVOLUTION  177 

3.  Expand  (3  m^  +  2  ri^)*. 

(3  m^  +  2  ny  =  [(3  m')  +  (2  n«)]^ 

=  (3  m^)^  +  4  (3  m')X2  7i')  +  6  (3  m^)^  (2  nS)^ 

+  4(3m2)(2n3)3+(2n3)4 
=  81  m«  +  216  mV  +  216  mV  +  96  mV  + 16  w^l 

A  polynomial  may  be  expressed  as  a  binomial,  and  raised  to 
any  positive  integral  power  by  successive  applications  of  the 
Binomial  Theorem. 

But  for  second  or  third  powers,  the  methods  of  §§  134  and 
136  are  shorter. 

4.  Expand  (a^  -  2  a;  -  2y, 

(x"  -2  x-2y=l(x^  -2  X)  -^  (-2)y 

=  (x'-2  xy-Jr  4(aj2  _  2  xfi -  2)  +  6(a^  -  2  xy(-  2y 
+  4.(x' -2  x)(-2y  +  (-2y 

=  x^-Sx'  +  24:a^-S2af  +  16x^ 

-S(x^-6x'  +  12x^-Sa^ 

+  24(a;*-4a^  +  4a;2)_32(a^_2a;)  +  16 

=  ic8_8fl;7_|_i6a;«  +  16aj^-56a^-32a^  +  64a^+64a;  +  16. 

287.    To  find  the  rth  or  general  term  in  the  expansion  of  (a  -1-  xy. 

The  following  laws  hold  for  any  term  in  the  expansion  of 
(a  +  a;)",  in  equation  (4),  §  285  : 

1.  The  exponent  of  x  is  less  by  1  than  the  number  of  the 
term. 

2.  The  exponent  of  a  is  n  minus  the  exponent  of  x. 

3.  The  last  factor  of  the  numerator  is  greater  by  1  than  the 
exponent  of  a. 

4.  The  last  factor  of  the  denominator  is  the  same  as  the 
exponent  of  x. 

Therefore,  in  the  ?'th  term,  the  exponent  of  x  will  be  r  —  1. 
The  exponent  of  a  will  be  n  —  (r  —  1),  or  w  —  r  -f  1. 
The  last  factor  of  the  numerator  will  be  n  —  r  +  2. 
The  last  factor  of  the  denominator  will  be  r  —  1. 


178  ADVANCED  COURSE  IN  ALGEBRA 

Hence,  the  rth  term 

^n(n-l)(n-2)".(n-r-\-2)  ^n-r+i^r-i  n^ 

1.2.3...(r-l)  *  ^^ 

In  finding  any  term  of  an  expansion,  it  is  convenient  to  obtain 
the  coefficient  and  exponents  of  the  terms  by  the  above  laws. 

Ex.     Find  the  8th  term  of  (3  a  -  b'y\ 

We  have,  (3  a  -  bY  =  [(3  a)  +  (-  6^)]". 
The  exponent  of  (—  ¥)  is  8  —  1,  or  7. 
The  exponent  of  (3  a)  is  11  —  7,  or  4. 

The  first  factor  of  the  numerator  is  11,  and  the  last  factor 
4  + 1,  or  5. 

The  last  factor  of  the  denominator  is  7. 

Then,  the  8th  term  ^  11  •  IQ  •  ^  •  8  »  7  ■  6  »  5  .3  ^yr^^y 
'  1.2.3.4.5.6.7   ^     ^  ^       ^ 

=  330  (81  a^)  (-  6^)  =  -  26730  a'b^. 

If  the  second  term  of  the  binomial  is  negative,  it  should  be  enclosed, 
sign  and  all,  in  parentheses  before  applying  the  laws. 

If  either  term  of  the  binomial  has  a  coefficient  or  exponent  other  than 
unity,  it  should  be  enclosed  in  parentheses  before  applying  the  laws. 

288.  Multiplying  both  terms  of  the  coefficient,  in  (1),  §  287, 
by  the  product  of  the  natural  numbers  from  1  to  n  —  r  +  1, 
inclusive,  the  coefficient  of  the  rth  term  becomes 

n(n  - 1)  . . .  (yi  -  r  +  2)  .  (n  -  r  + 1)  . . .  2  . 1  ^  |j^ 

I  r  — 1  xl  '2  ...(n  — r  +  1)  I  r  —  1  \  n  —  r-\-l 

To  find  the  coefficient  of  the  rth  term  from  the  end,  which 
since  the  number  of  terms  is  ti  + 1,  is  the  [n  —  (r  —  2)]th  from 
the  beginning,  we  put  in  the  above  formula  n  —  r-{-2  for  7\ 

Then,  the  coefficient  of  the  rth  term  from  the  end  is 

In  \  n 

I —  or 


I  n-r  +  2-1  I  n-(n-r  +  2)  +  l'        |  n-r-f  1  |  r-1 

Hence,  in  the  expansion  of  (a  +  xy,  the  coefficients  of  terms 
equidistant  from  the  ends  of  the  expansion  are  equal. 


INVOLUTION   AND  EVOLUTION  179 

289.  Properties  of  Coefficients  in  the  Expansion  of  (a  -f-  a;)**. 

I.  Putting  in  (4),  §  285,  a  =  l  and  a;  =  1,  we  have 

That  is,  tJie  sum  of  the  coefficients,  in  the  expansion  of  (a+ic)" 
is  equal  to  2**. 

II.  Putting  in  (4),  a  =  1  and  x  =  —  l,  we  have 

Q^l      y^  I  n(n-l)      n(n-l)(n-2)      _ 

[2         .  [3 

Or,       l  +  !%i)  +  ...  =  .  +  ^^"-|V"-^U-.       . 

That  is,  the  sum  of  the  coefficients  of  the  odd  terms  is  equal  to 
the  sum  of  the  coefficients  of  the  even  terms. 

290.  The  Greatest  Coefficient  in  the  Expansion  of  (a  +  a;)". 

By  §  287,  the  coefficient  of  the  (r  +  l)th  term,  in  the  expan- 
sion of  (a  +  ^)'N  is 

n(n  —  l)  ••'  (n  —  r-\-l) 
\r 

This  is  obtained  by  multiplying  the  coefficient  of  the  rth 

,      n  —  r-\-l         n-\-l      . 

term  by  ^^,  or  —I^- L 

r  r 

The  latter  expression  decreases  as  r  increases. 

It  is  evident  that  the  successive  coefficients,  commencing 

with  the  first  term,  will  increase  numerically  so  long  as 

■ IS  >  1. 

r 

I.    Suppose  n  even ;  and  let  n  =  2m,  where  m  is  a  positive 
integer. 

Then,   ^I — IIL-  becomes 

r  r 

If  r  =  m,  ^^~''  +  ^  becomes  '^  +  ^,  and  is  >  1. 
r  m 


180       ADVANCED  COURSE  IN  ALGEBRA 

U  r  =  m  +  l,  '^'^-r-Vl  becomes      ^     .  and  is  <  1. 
r  m  + 1 

Then,  the  greatest  coefficient  will  be  when  r  =  m-\- 1. 
As  the  number  of  terms  in  the  expansion  is  2  m  +  1,  it  fol- 
lows that  the  middle  term  has  the  greatest  coefficient. 

II.    Suppose  n  odd  ;  and  let  n  —  2m-\- 1,  where  m  is  a  posi- 
tive integer. 

Then, becomes  ^t_, 

r  r 

If  r  =  m,  'in  —  r-j-     ]3g(;omes  !??lir_,  and  is  >  1. 

r  m 

If  r  =  m  +  l,  — — ±—  becomes  ^  "^  ^  ,  and  equals  1. 

r  m-\-l 

T-p               ,o2m  —  r  +  2,                    m  -i-^h 

It  r  =  m-f  2,  ■ —  becomes  ,  and  is  <  1. 

'  r  m  +  2' 

There  will  then  be  two  terms  having  the  greatest  coefficient; 
those  where  r  =  m  +  1  and  r  =  m-{-2. 


EXERCISE   40 

Expand  the  following : 

1.  f-^J^y.  4.  (a +  6)6.  7.   (2a2_53)6. 

Expand  the  following  to  five  terms  : 

10.   (x  +  yy.  11.   (a6-l)io.  12.   (m2-2w3)ii. 

Expand  the  following : 

13.  (2  a;2  +  x  +  3)8.  16.  (2  ^2  +  x  -  4)*. 

14.  (4  a2  -  3  a6  -  62)8.  I7.  (i  +  2  a;  +  a;2)5. 

15.  (1 -3x+'2a;2)*.  18.  (?;'2  -  3  x  -  2)6. 

Find  the 

19.   8th  term  of  (a  +  x)".  20.   10th  term  of  (1  -  w)". 


INVOLUTION   AND   EVOLUTION  181 

21.  9tli  term  of  (a2  +  i)i5.  23.   7th  term  of  {x^  -  2  y^y\ 

22.  Sthtermof  f^-  — Y-  24.    Middle  term  of /'ma +  ^V^ 

25.  Term  involving  a;i4  in  (  a;* +  — |    • 

26.  Term  involving  x^i  in  (2x2 -)    . 

V  3xV 

EVOLUTION 

291.  Evolution  is  the  process  of  finding  any  root  (§  157)  of 
an  expression. 

We  shall  consider  in  the  present  chapter  those  cases  only  in  which  both 
the  expression  and  its  root  are  rational  (§  198). 

We  have  already  given  (§  166)  a  rule  for  finding  the  principal 
root  of  a  rational  and  integral  monomial,  which  is  a  perfect 
power  of  the  same  degree  as  the  index  of  the  required  root. 

292.  If  m  and  n  are  positive  integers,  we  have,  by  §  164, 

Whence,  by  §  157,       V(a^  =  ( Va")™. 

This  method  is  preferable  to  that  of  §  166,  if  the  expression 
whose  root  is  to  be  found  is  a  power  of  a  number  which  is  a  per- 
fect power  of  the  same  degree  as  the  index  of  the  root. 


Ex.    V(32  a}y  =  ( V32  a}^  =  (2  a?f  =  8  a\ 

293.   Any  Root  of  a  Fraction. 

Let  n  be  a  positive  integer,  and  a  and  h  numbers  which  are 
perfect  ?ith  powers. 

By  §  165,  ^^  X  ^b  =  y;lf^=  Va. 

Dividing  both  members  by  -^b, 
nja_;Va 


182  ADVANCED   COURSE  IN   ALGEBRA 

Then,  to  find  any  root  of  a  fraction,  each  of  whose  terms  is 
a  perfect  power  of  the  same  degree  as  the  index  of  the  required 
root,  extract  the  required  root  of  both  numerator  and  denominator, 
and  divide  the  first  result  by  the  second. 


Ex      i     27  ft-'^^^^^      -^27  a'b'^     3  ab^ 
^'       64  c»  -^64c"»  4c3' 


SQUARE  ROOT  OF  A  POLYNOMIAL 

294.  In  §  168,  we  gave  a  rule  for  finding  the  square  root  of 
a  trinomial  perfect  square ;  and,  in  §  170,  of  an  expression  of 
the  form  ^,  _^  b'-\-c'+2  ab-\-2ac  +  2  be, 

which  could  be  seen,  by  inspection,  to  be  a  perfect  square. 

We  will  now  consider  the  method  of  finding  the  square  root 
of  any  polynomial  perfect  square. 

Let  A  and  B  be  rational  and  integral  expressions  (§  63); 
and  suppose  them  to  be  arranged  in  the  same  order  of  powers 
of  some  common  letter,  x. 

Let  the  exponent  of  x  in  the  last  term  of  A  be  greater,  or 
less,  than  its  exponent  in  the  first  term  of  B,  according  as 
A  and  B  are  arranged  in  descending,  or  ascending,  powers 
of  X. 

By  §  131,  {A-irBf  =  A^  +  2AB  +  B', 

Whence,         {A  +  By-A'  =  2AB+  &. 

If  the  expression  2  AB  +  B^  be  arranged  in  the  same  order 
of  powers  of  a;  as  ^  and  B,  its  first  term  must  be  twice  the 
product  of  the  first  term  of  A  and  the  first  term  of  B. 

Hence,  the  first  term  of  B  may  be  obtained  by  dividing  the 
first  term  of  2  AB  -|-  B^  by  twice  the  first  term  of  A. 

By  the  expression  "first  term  of  ^,"  in  the  above  discussion,  we 
mean  the  sum  of  all  the  terms  of  A  containing  the  highest,  or  lowest, 
power  of  a;,  according  as  A  is  arranged  in  descending,  or  ascending, 
powers  of  as. 

Thus,  if  2I  =  ax*  +  6aj*  +  cx^,  the  first  terra  of  A  ia  (a  +  6) a*. 


INVOLUTION  AND   EVOLUTION  183 

A  similar  meaning  is  attached  to  the  expressions  "last  term  of  ^," 
and  "lirst  term  of  ^." 

295.   We  will  now  consider  an  example. 
Required  the  square  root  of 

24  a;  -  12  a.-^- 7  a^H- 4  a;*  +  16. 

Arranging  the  expression  according  to  the  descending  powers 
of  a?,  we  are  to  find  an  expression  which,  when  squared,  will 
produce  4  a;*  -  12  ar^  -  7  a:^  ^  24  aj  +  16. 

It  is  evident  from  §  134  that  the  first  term  of  the  expression 
is  the  square  of  the  term  containing  the  highest  power  of  x  in 
the  square  root. 

Hence,  the  term  containing  the  highest  power  of  x  in  the 
square  root  must  be  the  square  root  of  4  x'^,  or  2  a?. 

Denoting  the  term  of  the  root  already  found  by  A,  and  the 
remainder  of  the  root,  arranged  in  descending  powers  of  a;, 
by  B,  we  have 

(^-f--B)2-^2^4a;*-12a^-7a;2  +  24a;  +  16-(2a^2 

=  _12a^_7a^  +  24aj  +  16.  (1) 

By  §  294,  the  first  term  of  B  may  be  obtained  by  dividing 
the  first  term  of  (1),  —  12  a^,  by  twice  ^,  or  4  a^^j  that  is,  the 
first  term  of  5  is  —3x. 

Hence,  the  first  two  terms  of  the  root  are  2  a^  —  3  a;. 

Denoting  this  expression  by  A\  and  the  remainder  of  the 
root,  arranged  in  descending  powers  of  a?,  by  B\  we  have 

{A  +  By-A^'- 

=  4  a;^  - 12  «3 _  7aj2  + 24  ajH- 16- (2  a^- 3  a;)2 

=  4a;^-12aj«-7ar^-f24a;  +  16-(4a^-12a^  +  9a!2) 

=  _16a^4-24a;  +  16.  (2) 

By  §  294,  the  first  term  of  B'  may  be  obtained  by  dividing 
the  first  term  of  (2),  — 16  x^,  by  twice  the  first  term  of  A\ 
or  4  .T^ ;  that  is,  the  first  term  of  B'  is  —  4. 

Hence,  the  first  three  terms  of  the  root  are  2  ar^  —  3  a;  —  4. 


184 


ADVANCED  COURSE   IN   ALGEBRA 


Denoting  this  expression  by  A",  and  the  remainder  of  the 
root,  arranged  in  descending  powers  of  x,  by  B",  we  have 

(A"-^B"y-A"' 

=  4.x'-12a^-Tx^-\-24:X-{-16-(2a^-3x-4:y 

=4:x'-12x'~7x'+24.x+16-(4:x'-12x'~7x^-j-24.x-{-16) 

=0. 

Hence,  the  required  square  root  is  2  a^  —  3  a;  —  4. 

296.  Let  the  last  term  of  A'  be  O. 

Then,  A'  =  A+C,  2ind  A"  =  A' +  2AC  +  C\ 
Therefore, 

{A'  +  By  -  A^^  =  {A  +  Bf  -  A^  -2  AC  -  C^ 

=  \_{A^-By -  A'^-  {2  A+  C)a 
In  like  manner,  if  C"  denotes  the  last  term  of  A['j 

(^"  +  B"Y  -  A'"  =  [{A'  +  By  -  A"]  -(2A'-{-  G')C' ; 
and  so  on. 

That  is,  any  remainder  after  the  first  may  be  obtained  by 
subtracting  from  the  preceding  remainder  an  expression  which 
is  formed  by  doubling  the  part  of  the  root  already  found, 
adding  to  it  the  next  term  of  the  root,  and  multiplying  the 
result  by  this  term. 

The  expressions  2  A,  2  A',  etc.,  are  called  trial-divisors,  and  2  A  -{•  C, 
2A'+  C,  etc.,  complete  divisors. 

297.  It  is  customary  to  arrange  the  work  as  follows,  the  com- 
plete divisors  and  remainders  being  formed  by  the  rule  of  §  296  : 

4:x'-12x^-    7  a;^  + 24  a; +16 12x^-3  a;- 4 
4a;^ 


4a^ 


3a; 
3a; 


-12  a;^-   7  a;2 -f  24  a;  +  16,  1st  Eem. 
-12a.'3+    90^ 


4ar^ 


6x  —   4 
-   4 


- 16  x2  -I-  24  a;  + 16,  2d  Kern. 
-16x2  +  24  a;  +  16 


To  avoid  needless  repetition,  the  last  three  terms  of  the  first  remainderj 
and  the  last  two  terms  of  the  second,  may  be  omitted. 


INVOLUTION   AND   EVOLUTION  185 

We  then  have  the  following  rule  for  extracting  the  square 
root  of  a  polynomial  perfect  square : 

Arrange  the  expression  according  to  the  powers  of  some  letter. 

Extract  the  square  root  of  the  first  term{%  294),  write  the  result 
as  the  first  term  of  the  root,  and  subtract  its  square  from  the  given 
expressioyi,  arranging  the  remainder  in  the  same  order  of  powers 
as  the  given  expression. 

Divide  the  first  term  of  the  remainder  by  twice  the  first  term  of 
the  root,  and  add  the  quotient  to  the  part  of  the  root  already  found, 
and  also  to  the  trial-divisor. 

Multiply  the  complete  divisor  by  the  term  of  the  root  last  obtained, 
and  subtract  the  product  from  the  remainder. 

If  other  terms  remain,  proceed  as  before,  doubling  the  part  of 
the  root  already  found  for  the  next  trial-divisor. 

If  the  expression  had  been  written 

16  +  24  a;  -  7  x2  -  12  x3  +  4  a:*, 

the  square  root  would  have  been  obtained  in  the  form  4  +  3  x  —  2  cc^, 
which  is  the  negative  of  2  aj2  _  3  gj  _  4. 

298.  With  the  notation  of  §  296, 

2^'  =  2^  +  2a 
In  like  manner,  2  A^'  =  2  A'  -\-2  C ;  and  so  on. 

That  is,  any  trial-divisor,  after  the  first,  is  equal  to  the  preced- 
ing complete  divisor  with  its  last  term  doubled. 

SQUARE  ROOT  OF  AN  ARITHMETICAL  NUMBER 

The  term  '■^  number,^''  in  the  following  discussion,  signifies  an  integral 
or  decimal  perfect  square,  expressed  in  Arabic  numerals. 

299.  The  square  root  of  100  is  10 ;  of  10000  is  100 ;  etc. 
Hence,  the  square  root  of  a  number  between  1  and  100  is 

between  1  and  10;  the  square  root  of  a  number  between  100 
and  10000  is  between  10  and  100 ;  etc. 

That  is,  the  square  root  of  an  integer  of  one  or  two  digits 
contains  one  digit;  the  square  root  of  an  integer  of  three  or 
four  digits  contains  two  digits ;  etc. 


186  ADVANCED   COURSE   IN   ALGEBRA 

Hence,  if  a  point  he  placed  over  every  second  digit  of  an  integer, 
beginning  at  the  units'  place,  the  number  of  points  shows  the  number 
of  digits  in  its  square  root. 

300.  If  a  is  an  integral  perfect  square,  then  -— ,  where  n  is 
any  positive  integer,  is  also  a  perfect  square. 

But  is  a  number  whose  decimal  part  contains  an  even 

10-" 

number  of  digits,  and  which  differs  from  a  only  in  the  position 
of  its  decimal  point. 

Hence,  if  a  point  be  placed  over  every  second  digit  of  any 
number,  beginning   at  the  units'  place  and  extending  in  either 
direction,  the  yiumber  of  points  shows  the  yiumber  of  digits  in  its  I 
square  root. 

301.  Let  a,  b,  and  c  represent  positive  integers. 
We  have, 

(a  +  5  +  c)^-a^^2a(&  +  c)4-(&  +  c)^^^      ^     {b^-c)\ 
2a  2a  2a 

That  is,  if  the  remainder  obtained  by  subtracting  a^  from 
(a  +  6  +  cy  be  divided  by  2  a,  the  quotient  is  greater  than  b. 

In  like  manner,  if  the  remainder  obtained  by  subtracting  a^ 
from  (a -{-by  be  divided  by  2  a,  the  quotient  is  greater  than  b. 

302.  We  will  now  consider  an  example. 
Eequired  the  square  root  of  10719076. 

Pointing  the  number  in  accordance  with  the  rule  of  §  299, 
we  find  that  there  are  four  digits  in  its  square  root. 

Since  the  number  is  between  9000000  and  16000000,  the 
square  root  is  between  3000  and  4000. 

That  is,  the  first  digit  of  the  root  is  3. 

Let  a  represent  the  number  3000 ;  b  the  second  digit  of  the 
root,  multiplied  by  100 ;  and  c  the  number  consisting  of  the 
last  two  digits  of  the  root  in  their  order. 

Then,  a  +  &  -f  c  represents  the  root ;  now, 

(a^b-{-cy-a'^  10719076  - 9000000 _  1719076  _  ^^^  ^ 
2  a  6000  6000 


INVOLUTION   AND   EVOLUTION  187 

By  §  301,  this  is  greater  than  h. 

Hence,  6  is  a  multiple  of  100  less  than  286.  +  . 

Assume,  then,  h  =  200. 

Then,  the  first  two  digits  of  the  root  would  be  32. 

Let  a'  represent  the  number  3200 ;  h'  the  third  digit  of  the 
root,  multiplied  by  10 ;  and  c'  the  last  digit  of  the  root. 

Then,  a^  -\-h^  -{-c'  represents  the  root ;  now, 

(g^  4-  6^  4.  cy  _  a'2  ^  10719076-10240000  ^  479076  ^  n>^  . 
2  a'  6400  6400  '    ' 

By  §  301,  this  is  greater  than  5'. 

Hence,  V  is  a  multiple  of  10  less  than  74.  +  . 

Assume,  then,  6' =  70. 

Then,  the  first  three  digits  of  the  root  would  be  327. 

Let  a"  represent  the  number  3270,  and  6"  the  last  digit  of 
the  root. 

Then,  a"  +  6"  represents  the  root ;  now, 

(gn  -4-  6>y  -  a^'2  ^  10719076  _  10692900  ^  26176  ^  ^  _^ 
2  a"  6540  6540         '    ' 

By  §  301,  this  is  greater  than  6"  ;  assume,  then,  &"  =  4. 

Since  (3274)2  =  10719076,  the  required  square  root  is  3274. 

303.   We  have  with  the  notation  of  §  302, 

(g'  +  5'_l_c')2_a'2=(a+&+-c)2-(a  +  6)2 

z=  {a-\-h  +  cf  -  a?  -2  ah  -W 

=  [(a  +  6  +  cf  -  a^]  -  (2  a  +  h)h. 
Similarly, 

(g"  +  6")2  _  g"2  ^  [-(^r  _^  5f  _^  ^y  _  ^f2-|  _  (2  g'  +  &')6'. 

That  is,  any  remainder  after  the  first  may  be  obtained  by 
subtracting  from  the  preceding  remainder  a  number  which 
is  formed  by  doubling  the  part  of  the  root  already  obtained, 
adding  to  it  the  next  root  digit  followed  by  as  many  ciphers  as 
there  are  digits  in  the  remainder  of  the  root,  and  multiplying 
the  result  by  the  latter  number. 

The  numbers  represented  by  .2  a,  2  a',  etc.,  are  called  trial-divisors^ 
and  those  represented  by  2  a  +  6,  2  a'  +  &',  etc.,  complete  divisors. 


188 


ADVANCED  COURSE  IN  ALGEBRA 


304.  The  work  of  the  example  of  §  302  may  be  arranged 
as  follows,  the  complete  divisors  and  remainders  being  formed 
by  the  rule  of  §  303  : 

16719076  I  3000  +  200  +  70+4 

a^- 9000000 

1st  Comp.  Div., 

2d  Comp.  Div., 

3d  Comp.  Div., 


Omitting  the  ciphers  for  the  sake  of  brevity,  and  condensing 
the  operation,  it  will  stand  as  follows  : 


6000  +  200 
200 

1719076 
1240000 

6400  +  70 

70 

479076 
452900 

6540+4 
4 

26176 
26176 

10719076 
9 


3274 


62 

171 
124 

647 

4790 
4529 

6544 

[   26176 
26176 

We  then  have  the  following  rule  for  extracting  the  square 
root  of  an  integral  perfect  square  : 

Separate  the  number  into  periods  by  pointing  every  secoyid  digit, 
beginning  with  the  units'  place. 

Find  the  greatest  square  in  the  left-hand  period,  and.  write  its 
square  root  as  the  first  digit  of  the  root;  sxibtract  the  squai'e  of  the 
first  root-digit  from  the  left-hand  period,  and  to  the  result  annex 
the  next  period. 

Divide  this  remainder,  omitting  the  last  digit,  by  ttvice  the  jxirt 
of  the  root  already  found,  and  annex  the  quotient  to  the  root,  and 
also  to  the  trial-divisor. 

Multiply  the  complete  divisor  by  the  root-digit  last  obtained,  and 
subtract  the  product  from  the  remainder. 


INVOLUTION   AND  EVOLUTION  189 

If  other  periods  remain,  proceed  as  before,  doiibling  the  part 
of  the  root  already  found  for  the  next  trial-divisor. 

Note  1.  It  sometimes  happens  that,  on  multiplying  a  complete  divisor 
by  the  digit  of  the  root  last  obtained,  the  product  is  greater  than  the 
remainder.  In  such  a  case,  the  digit  of  the  root  last  obtained  is  too  great, 
and  one  less  must  be  substituted  for  it. 

Note  2.  If  any  root-digit  is  0,  annex  0  to  the  trial-divisor,  and  annex 
to  the  remainder  the  next  period. 

305.  We  will  now  show  how  to  obtain  the  square  root  of  a 
number  which  is  not  integral. 

Eequired  the  square  root  of  49.449024. 

Wehave,  V4al4902i  =  ^,'^g  =  I5| (§  293)  =  7.032. 

The  work  may  be  arranged  as  follows  : 

49.44902417.032 
49 


1403 

4490 
4209 

14062 

28124 
28124 

Since  14  is  not  contained  in  4,  we  write  0  as  the  second  root-digit,  in 
the  above  example  ;  we  then  annex  0  to  the  trial-divisor  14,  and  annex  to 
the  remainder  the  next  period,  90. 

Hence,  if  any  number  be  pointed  in  accordance  with  the  rule 
of  §  300,  the  rule  of  §  304  may  be  applied  to  the  result,  and 
the  decimal  point  inserted  in  its  proper  position  in  the  root. 

306.  After  n-\-l  digits  of  the  square  root  of  an  integral 
perfect  square  have  been  found  by  the  rule  of  §  304,  n  more 
may  be  obtained  by  simple  division  only,  supposing  2n-\-Xto 
be  the  whole  number. 

For  let  a  represent  the  integer  whose  first  n-\-l  digits  are 
the  first  w  -f-  1  digits  of  the  root  in  their  order,  and  whose  last 
n  digits  are  ciphers;  and  let  h  represent  the  integer  consisting 
of  the  last  n  digits  of  the  root  in  their  order. 

Then,  a-\-h  represents  the  root. 


190  ADVANCED   COURSE   IN   ALGEBRA 


We  have, 

(a 

+  by- 

-a^  = 

2ab-{-  h\ 

Whence, 

(a 

+  by- 

2a 

-a' 

'-fe 

That  is,  (a 

-\-by- 

-  a^,  divided 

by  2  a, 

digits  of  the  root,  increased  by  - — 

52  ^ 

We  will  now  prove  that  -—  is  less  than  -  ;  so  that,  by  neg- 
Z  a  Li 

lecting  the  remainder  arising  from  the  division,  we  obtain  the 
part  of  the  root  required. 

By  hypothesis,  h  contains  n  digits. 

Then,  6^  cannot  contain  more  than  2  n  digits. 

But  a  contains  2  n  +  1  digits.     • 

52  6^  1 

Hence,  -  is  less  than  1 :  and  therefore  —  is  less  than  — 

a  2a  .  2 

If,  then,  the  (n  +  l)th  remainder  be  divided  by  twice  the 
part  of  the  root  already  found,  the  remaining  n  digits  of  the 
root  may  be  obtained. 

The  method  applies  to  the  square  root  of  any  number. 

Ex.    Required  the  square  root  of  638.876176. 

638.876i76  [25^ 
4 


45 

238 
225 

502 

1387 
1004 

50.4)3.836176(.076 
3528 
3081 

We  obtain  the  first  three  digits  of  the  root  by  the  ordinary 
method,  and  the  other  two  by  the  method  of  §  306;  that  is, 
by  dividing  the  third  remainder,  3.836176,  by  twice  the  part  of 
the  root  already  obtained,  or  50.4. 

The  required  root  is  25.2  +  .076,  or  25.276. 


INVOLUTION   AND   EVOLUTION  191 

CUBE  ROOT   OF  A  POLYNOMIAL 

307.  In  §  176,  we  gave  a  method  for  finding  the  cube  root 
of  any  expression  of  the  form 

a^  H-  3  a^fe  +  3  ab^  +  b%  or  a^  -  3  a'b  -\-Sab^-  b\ 

We  will  now  consider  the  method  of  finding  the  cube  root  of 
any  polynomial  perfect  cube. 

Let  A  and  B  have  the  same  meanings  as  in  §  294. 

By  §  135,  {A  +  B)3  =  ^^  +  3  A^B  +  3  AB'  -f  B\ 

Whence,  {A  +  Bf  -A^  =  ?>  A'B  +  3  AB'  +  B\ 

If  the  expression  3  A^B  +  3  AB~  +  B^  be  arranged  in  the  same 
order  of  powers  of  x  as  A  and  B,  its  first  term  must  be  three 
times  the  product  of  the  square  of  the  first  term  of  A  and  the 
first  term  of  B. 

Hence,  the  first  term  of  B  may  be  obtained  by  dividing  the 
first  term  of  3  A^B  +  3  AB^  +  B^  by  three  times  the  square  of 
the  first  term  of  A. 

308.  We  will  now  consider  an  example. 
Kequired  the  cube  root  of 

A0a^-6a:^-64.  +  x^-96x. 

Arranging  the  expression  according  to  the  descending  powers 
of  X,  we  are  to  find  an  expression  which,  when  cubed,  will 
produce  x^ _  6  af  +  ^0  x" -96  x -  64:. 

It  is  evident  from  §  136  that  the  first  term  of  the  expression 
is  the  cube  of  the  term  containing  the  highest  power  of  x  in  the 
cube  root. 

Hence,  the  term  containing  the  highest  power  of  x  in  the  cube 
root  must  be  the  cube  root  of  a?^,  or  a^. 

Denoting  the  term  of  the  root  already  found  by  A,  and  the 
remainder  of  the  root,  arranged  in  descending  powers  of  x,  by 
B,  we  have 

(A  +  Bf-A^  =  af-6  3^  +  i0x'-96x-6i-(xJ 

=  _  6  a^  4-  40  ar^  -  96  x  -  64.  (1) 


192  ADVANCED   COURSE   IN   ALGEBRA 

By  §  307,  the  first  term  of  B  may  be  obtained  by  dividing 

the  first  term  of  (1),  —  6  x',  by  three  times  the  square  of  A,  or 

3  x"^ ;  that  is,  the  first  term  of  jB  is  —  2  x. 

Hence,  the  first  two  terms  of  the  root  are  x^  —  2x. 
Denoting  this  expression  by  A',  and  the  renjainder  of  the 

root,  arranged  in  descending  powers  of  x,  by  B',  we  have 

(A'  +  B'f-A" 

=  a;«  -  6  a^  +  40  a^  -  96  a;  -  64  -  (a^  -  2  ar) 3 

=  a;«  _  6  a^  +  40  a^  -  96  a;  -  64  -  (a;*5  -  6  a;^  + 12  a;^  -  8  a;3) 

=  _12aj4_|.4g^_9g^_64  (2) 

Then,  the  first  term  of  B'  may  be  obtained  by  dividing  the 
first  term  of  (2),  —  12  a;'*,  by  three  times  the  square  of  the  first 
term  of  A',  or  3  aj"* ;  that  is,  the  first  term  of  B'  is  —  4. 

Hence,  the  first  three  terms  of  the  root  are  x^  —  2  x  —  4. 

Denoting  this  expression  by  A",  and  the  remainder  of  the 
root,  arranged  in  descending  powers  of  x,  by  B",  we  have 
(A"  +  By  -  A"' 

=  a;6  -  6  a^  4-  40  a^  -  96  a;  -  64  -  (a;2  -  2  a;  -  4)3  =  0. 

Hence,  the  required  cube  root  is  »^  —  2  a;  —  4. 

309.   Let  the  last  term  of  A'  be  C. 
Then,  A'  =  A -{-  C;  whence, 
(A'  +  By  -  A''  =(A-\-By-A^-S  A'C  -  3  AC  -  C^ 

=  l(A  +  By -A'-]  ~(SA'-\-3AC+  C')G. 
In  like  manner,  if  C"  denotes  the  last  term  of  A'\ 

and  so  on. 

That  is,  any  remainder  after  the  first  may  be  obtained  by 
subtracting  from  the  preceding  remainder  an  expression  which 
is  formed  by  adding  together  three  times  the  square  of  the 
part  of  the  root  already  found,  three  times  the  product  of  the 
part  of  the  root  already  found  by  the  next  term  of  the  root, 
and  the  square  of  the  next  term  of  the  root,  and  multiplying 
the  sum  by  the  latter  term. 


INVOLUTION   AND   EVOLUTION  193 

The  expressions  SA^,  SA''^,  etc.,  are  called  trial-divisors,  and  3J.2  + 
SAC+  C^SA'^^  +  SA'C  +  C'-^,  etc.,  complete  divisors. 

310.   We  arrange  the  work  as  follows,  the  complete  divisors 
and  the  remainders  being  formed  by  the  rule  of  §  309 : 


x^-6Q^-\-^0a^-9Qx  -64 
x' 


x'-2x-4: 


3a;4_6ar^  +  4a^ 


-6a^-^^0x^-96x  -64 

-6a^-^12x'-    Sa^ 

Sx'-naf  +  Ux'  -12x''-\-4.Sx^-96x  -64 

-12ar^  +  24a;  +  16 

3a;*-12af^  +24a;  +  16|  -  12a;^ +  48a^  -  96a;  -64 

The  last  three  terms  of  the  first  remainder,  and  the  last  two  terms  of 
the  second,  may  be  omitted. 

We  then  have  the  following  rule  for  extracting  the  cube  root 
of  a  polynomial  perfect  cube  : 

Arrange  the  expression  according  to  the  powers  of  some  letter. 

Extract  the  cube  root  of  the  first  term,  icrite  the  result  as  the 
first  term  of  the  root,  and  subtract  its  cube  from  the  given  expres- 
sion ;  arranging  the  remainder  in  the  same  order  of  powers  as  the 
given  expression. 

Divide  the  first  term  of  the  remainder  by  three  times  the  square 
of  the  first  term  of  the  root,  and  write  the  residt  as  the  next  term 
of  the  root.  ■       - 

Add  to  the  trial-divisor  three  times  the  product  of  the  term  of 
the  root  last  obtained  by  the  part  of  the  root  previously  found,  and 
the  square  of  the  term  of  the  root  last  obtained. 

Multiply  the  complete  divisor  by  the  term  of  the  root  last  ob- 
tained, and  subtract  the  result  from  the  remainder. 

If  other  terms  remain,  proceed  as  before,  taking  three  times  the 
square  of  the  part  of  the  root  already  found  for  the  next  trial- 
divisor. 

311.   With  the  notation  of  §  309, 

3  ^'2  =  3  (^  +  C)2  =  3  ^2  +  6  ^(7  +  30^ 
=  3  ^2 _^ 3  ^(7-hC2+ (3  JIC +  2  02). 


194  ADVANCED  COURSE   IN   ALGEBRA 

In  like  manner, 

3  A'"  =  3  A" +  3  A'C'-\-  C"  +  (3  A'CT  +  2  C")  ;  etc. 

Tliat  is,  if  the  last  term  of  the  expression  which  is  added  to  any 
trial-divisor  he  doubled,  the  residt,  added  to  the  corresponding 
complete  divisor,  will  give  the  next  trial-divisor. 

Thus,  in  the  example  of  §  310,  if  we  add  to  the  first  complete  divisor 
Sx*-Qx^  +  4:  x^^the  expression  -6x^-\-Sx^,  the  result,  3  x*  -  12  x^  + 12  x^, 
is  the  next  trial-divisor. 

CUBE  ROOT  OF  AN  ARITHMETICAL  NUMBER 

The  term  '■'■  ymmher^''''  in  the  following  discussion,  signifies  a  positive 
integral  or  decimal  perfect  cube,  expressed  in  Arabic  numerals. 

312.  The  cube  root  of  1000  is  10;  of  1000000  is  100;  etc. 

Hence,  the  cube  root  of  a  number  between  1  and  1000  is  be- 
tween 1  and  10 ;  the  cube  root  of  a  number  between  1000  and 
1000000  is  between  10  and  100 ;  etc. 

That  is,  the  cube  root  of  an  integer  of  one,  two,  or  three 
digits  contains  one  digit ;  the  cube  root  of  an  integer  of  four, 
five,  or  six  digits  contains  two  digits ;  etc. 

Hence,  if  a.  point  he  placed  over  every  third  digit  of  an  integer, 
heginning  at  the  units''  jolace,  the  numher  of  points  shows  the  num- 
ber of  digits  in  its  cube  root. 

313.  If  a  is  an  integral  perfect  cube,  then  —j^,  where  w  is  any 
positive  integer,  is  also  a  perfect  cube. 

But  T-—  is  a  number,  the  number  of  digits  in  whose  decimal 

part  is  divisible  by  3,  and  which  differs  from  a  only  in  the 
position  of  its  decimal  point. 

Therefore,  if  a  point  be  placed  over  every  third  digit  of  any 
numher,  heginning  at  the  units''  place  and  extending  in  either 
direction,  the  number  of  points  shows  the  number  of  digits  in  its 
cube  root. 

314.  Let  a,  b,  and  c  represent  positive  integers. 


INVOLUTION   AND   EVOLUTION  195 

Then    (a  +  &  +  cy-ft^^3a^(6  +  c)  +  3a(&H-cy  +  (&  +  c)» 

^^  ^  ^  ,  3a(b  +  cy  +  (b  +  cY 
3  a- 
That  is,  if  the  remainder  obtained  by  subtracting  a^  from 
(a  +  6  +  cy  be  divided  by  3  a^,  the  quotient  is  greater  than  b. 

Similarly,  if  the  remainder  obtained  by  subtracting  a^  from 
(a  +  by  be  divided  by  3  a^,  the  quotient  is  greater  than  b. 

315.    We  will  now  consider  an  example. 

Required  the  cube  root  of  9745491456. 

Pointing  the  number  in  accordance  with  the  rule  of  §  312, 
we  find  that  there  are  four  digits  in  its  cube  root. 

Since  the  number  is  between  8000000000  and  27000000000, 
the  cube  root  is  between  2000  and  3000. 

That  is,  the  first  digit  of  the  root  is  2. 

Let  a  represent  the  number  2000 ;  b  the  second  digit  of  the 
root,  multiplied  by  100 ;  and  c  the  number  consisting  of  the 
last  two  digits  of  the  root  in  their  order. 

Then,  a-{-b  +  c  represents  the  root ;  now, 

(a^b-\-  cy  -  g-^  ^  9745491456  -  8000000000 
So"  12000000 

^  1745491456  _-^^g^ 
12000000 
By  §  314,  this  is  greater  than  b. 
Hence,  6  is  a  multiple  of  100  less  than  145. +. 
Assume,  then,  b  =  100. 

Then,  the  first  two  digits  of  the  root  would  be  21. 
Let  a'  represent  the  number  2100;  b'  the  third  digit  of  the 
root  multiplied  by  10;  and  c'  the  last  digit  of  the  root. 
Then,  a'  -\-b'  -\-  c'  represents  the  root ;  now, 

(a'  -hb'  +  c'y  -  a"  ^  9745491456  -  9261000000 
3  a'2  13230000 

484491456  ^3g^ 


13230000 


196  ADVANCED   COURSE   IN   ALGEBRA 

By  §  314,  this  is  greater  than  b'. 
Hence,  b'  is  a  multiple  of  10  less  than  36.  +  . 
Assume,  then,  b'  =!=  30. 

Then,  the  first  three  digits  of  the  root  would  be  213. 
Let  a"  represent  the  number  2130,  and  b"  the  last  digit  of 
the  root. 

Then,  a"  +  b"  represents  the  root;  now, 

(a"  +  b"y  -  a"^  ^  9745491456  -  9663597000 
3a"2  13610700 

^  81894456  ^  ^  _^ 
13610700        ■    * 

By  §  314,  this  is  greater  than  b" ;  assume,  then,  b"  =  6. 
Then,  since  (2136)^  =  9745491456,  the  required  cube  root  is 
2136. 

316.  We  have  with  the  notation  of  §  315, 
(a'  +  &'  +  Cy  -a'^=(a  +  b  +  cf  -  (a  +  bf 

=  (a  +  b-{-cy-a^-Sa'b-3ab^-b^ 

=  [  (a  +  &  +  c)3  -  a^]  -  (3  a^  +  3  a6  +  &-)  6. 
Similarly, 

(a"  +  by-a"^  =  [(a'^+  b'  +  cy-a'']-(3a"-\-S  a'b'-{-b")b'. 

That  is,  any  remainder  after  the  first  may  be  obtained  by 
subtracting  from  the  preceding  remainder  a  number  which  is 
formed  by  taking  three  times  the  square  of  the  part  of  the  root 
already  obtained,  adding  to  it  three  times  the  product  of  the 
part  of  the  root  already  obtained  by  the  next  root-digit  fol- 
lowed by  as  many  ciphers  as  there  are  digits  in  the  remainder 
of  the  root,  plus  the  square  of  the  latter  number,  and  multiply- 
ing the  result  by  the  latter  number. 

The  numbers  represented  by^  3  a^,  3  a'^,  etc. ,  are  called  trial-divisors, 
and  those  represented  by  Sa^  +  3ab  +  b%  3  a'^  +  3  a'b'  +  6'^,  etc.,  com- 
plete divisors. 

317.  The  work  of  the  example  of  §  315  may  be  arranged  as 
follows,  the  complete  divisors  and  remainders  being  formed  by 
the  rule  of  §  316. 


INVOLUTION   AND   EVOLUTION 


197 


9745491456  |  2000  +  100  +  30  +  6 

a^  =  8000000000 

12000000 

1745491456 

600000 

10000 

1st.  Comp.  Div., 

12610000 

1261000000 

13230000 

484491456 

189000 

900 

2d  Comp.  Div., 

13419900 
13610700 

402597000 

81894456 

38340 

36 

3d  Comp.  Div., 

13649076 

81894456 

Condensing  the  operation,  it  will  stand  as  follows 

974549i456  |  2136 
8 


1200  : 

60 

1 

L745 

1261  : 

L261 

132300 

1890 

9 

484491 

134199 

402597 

13610700 

38340 

36 

81894456 

13649076 

81894456 

We  then  have  the  following  rule  for  extracting  the  cube  root 
of  an  integral  perfect  cube : 

Separate  the  number  into  periods  by  pointing  every  third  digit y 
beginning  with  the  units'  place. 


198       ADVANCED  COURSE  IN  ALGEBRA 

Find  the  greatest  cube  in  the  left-hand  period,  and  write  its  cube 
root  as  the  first  digit  of  the  root;  subtract  the  cube  of  the  first  root- 
digit  from  the  left-hand  period,  and  to  the  result  aimex  the  next 
period. 

Divide  this  remainder  by  three  times  the  square  of  the  part  of 
the  root  already  found,  with  two  ciphers  annexed,  and  write  the 
quotient  as  the  next  digit  of  the  7'oot. 

Add  to  the  trial-divisor  three  times  the  product  of  the  last  root- 
digit  by  the  part  of  the  root  previously  found,  with  one  cipher 
annexed,  and  the  square  of  the  last  root-digit. 

Multiply  the  complete  divisor  by  the  digit  of  the  root  last 
obtained,  and  subtract  the  product  from  the  reynainder. 

If  other  periods  remain,  proceed  as  before,  takiiig  three  times 
the  square  of  the  part  of  the  root  already  found,  with  two  ciphers 
annexed,  for  the  next  trial-divisor. 

Note  1,  §  304,  applies  with  equal  force  to  the  above  rule. 

If  any  root-digit  is  0,  annex  two  ciphers  to  the  trial-divisor,  and  annex 
to  the  remainder  the  next  period. 

318.  With  the  notation  of  §  315, 

3  a'2  =  3  (a  -f  bf  =  3  a^  +  6  a&  +  3  6^ 
=  3  a^  -f-  3  a6  +  6M-  (3  ab  +  2  b^). 
In  like  manner, 

3  a"2  =  3  a'2  +  3  a'V  -f  b'^  +  (3  a'Z>'  -f  2  V') ;  etc. 
That  is,  if  the  first  number  and  the  double  of  the  second  number 
required  to  complete  any  trial-divisor  be  added  to  the  complete 
divisor,  the  residt,  tvith  tivo  ciphers  annexed,  will  give  the  next 
trial-divisor. 

319.  We  will  now  show  how  to  obtain  the  cube  root  of  a 
number  which  is  not  integral. 

Kequired  the  cube  root  of  1073.741824. 

We.ave,     ^1^21  =  4^^ 

=  12?i(§  293)  =  10.24. 
100   ^         ^ 

The  work  may  be  arranged  as  follows: 


INVOLUTION   AND    EVOLUTION  199 


1073.741824  10.24 

1 

30000 

73741 

600 

4 

30604 

61208 

600 

12533824 

8 

3121200 

12240 

16 

3133456 

12533824 

Here  the  second  root-digit  is  0 ;  we  then '  annex  two  ciphers  to  the 
trial-divisor  300,  and  annex  to  the  remainder  the  next  period,  741. 

The  second  trial-divisor  is  formed  by  the  rule  of  §  318. 

Adding  to  the  complete  divisor  30604  the  first  number,  600,  and  twice 
the  second  number,  8,  required  to  complete  the  trial-divisor  30000,  we 
have  31212  ;  annexing  two  ciphers  to  this,  the  result  is  3121200. 

Hence,  if  any  number  be  pointed  in  accordance  with  the  rule 
of  §  313,  the  rule  of  §  317  may  be  applied  to  the  result,  and 
the  decimal  point  inserted  in  its  proper  position  in  the  root. 

320.  After  n  -{- 2  digits  of  the  cube  root  of  an  integral  per- 
fect cube  have  been  found  by  the  rule  of  §  317,  n  more  may  be 
obtained  by  division,  supposing  2  n  -f  2  to  be  the  whole  number. 

For  let  a  represent  the  integer  whose  first  w  +  2  digits  are 
the  first  n  -J-  2  digits  of  the  root  in  their  order,  and  whose  last 
n  digits  are  ciphers,  and  b  the  integer  consisting  of  the  last  n 
digits  of  the  root  in  their  order ;  then,  a+6  represents  the  root. 

We  have,        (a  +  by-a^=^3  a'b-\-3  aW  +  b\ 

Whence,         ^02 =  ^^ ^■^"i* 

3  a^  a      3a^ 

That  is,  (a-\-bf  —  a^,  divided  by  3  a^,  will  give  the  last  n 

52       ^3 
digits  of  the  root,  increased  by  —  +  ;;—;• 

a      3a^ 

By  hypothesis,  b  contains  n  digits. 

Then,  b^  cannot  contain  more  than  2  n  digits; 


200 


ADVANCED  COURSE  IN  ALGEBRA 


But  a  contains  2  n  +  2  digits 
Again, 


52  H 

and  hence  —  is  less  than  —  • 
a  10 


— -  =  —  X  TT— ;  and  since  —  is  less  than  — - ,  and  - — 
oa^      a      o  a  a  10  oa 


53  1 

less  than  1,  — -  is  also  less  than  — • 
3  ci^  10 

WW  1  ' 

Therefore,  —  +  -—  is  less  than  - . 
a      3a^  .       ^ 

If,  then,  the  (n  +  2)th  remainder  be  divided  by  three  times 
the  square  of  the  part  of  the  root  already  found,  the  remaining 
n  digits  of  the  root  may  be  obtained. 

The  method  applies  to  the  cube  root  of  any  number. 

Ex.    Required  the  cube  root  of  1452648.865311064. 


1452648.865311064 
1 


113.2 


300 

452 

30 

1 

331 

331 

30 

121648 

2 

36300 

990 

9 

37299 

111897 

990 

9751865 

18 

3830700 

6780 

4 

383748 

4 

7674968 

6780 
8 


38442.72)2076.897311064(.054 
19221360 
15476131 


INVOLUTION   AND  EVOLUTION  201 

We  obtain  the  first  four  digits  of  the  root  by  the  ordinary 
method,  and  the  other  two  by  §  320;  that  is,  by  dividing  the 
fourth  remainder,  2076.897311064,  by  three  times  the  square 
of  the  part  of  the  root  already  found,  or  38442.72. 

The  required  root  is  113.2  +  .054,  or  113.254. 

ANY  ROOT  OF  A  POLYNOMIAL 

321.  Let  A  and  B  have  the  same  meaning  as  in  §  294. 
By  §  285,  if  n  is  any  positive  integer, 

(A  +  Bf  =  J."  +  nA^'-'^B  H . 

Whence,      (A  +  By  -  ^"  =  nA^'-^B  +  •  •  •• 

If  the  expression  nA'^~'^B-\-  •••he  arranged  in  the  same  order 
of  powers  of  a;  as  ^  and  B,  its  first  term  must  be  n  times  the 
product  of  the  (n  —  l)th  power  of  the  first  term  of  A  by  the 
first  term  of  B. 

Hence,  the  first  term  of  B  may  be  obtained  by  dividing  the 
first  term  of  the  expression  nA'^^^B  +  -"by  w  times  the  (n— l)th 
power  of  the  first  term  of  A. 

322.  It  follows  from  §  321,  exactly  as  in  §§  295  and  308, 
that  the  nth  root  of  a  polynomial,  which  is  a  perfect  power  of 
the  nth  degree,  may  be  found  by  the  following  rule : 

Arrange  the  expression  according  to  the  powers  of  some  letter. 

Extract  the  nth  root  of  the  first  teryn,  and  write  the  result  as  the 
first  term  of  the  root;  subtract  f^om  the  polynomial  its  first  term, 
and  arrange  the  remainder  in  the  same  order  of  powers  as  the 
given  expression. 

Divide  the  first  term  of  the  remainder  by  n  times  the  (n  —  l)th 
power  of  the  first  term  of  the  root,  and  write  the  result  as  the 
second  term  of  the  root. 

Subtract  from  the  given  polynomial  the  nth  power  of  the  part  of 
the  root  already  fouyid,  and  arrange  the  remainder  in  the  same 
order  of  powers  as  the  given  expression. 

If  other  terms  remain,  proceed  as  before,  dividing  the  first  term 
of  the  remainder  by  n  times  the  (n  —  l)th  power  of  the  first  term 
of  the  root;  arid  continue  in  this  manner  until  there  is  no 
remainder. 


202  ADVANCED   COURSE   IN   ALGEBRA 

ANY  ROOT  OF  AN  ARITHMETICAL  NUMBER 

The  term  "  number^''''  in  the  following  discussion,  signifies  a  positive, 
integral  or  decimal,  perfect  power  of  the  degree  denoted  by  the  index  of 
the  required  root,  expressed  in  Arabic  numerals. 

323.  It  may  be  proved,  as  in  §§  299,  300,  312,  and  313,  that : 
If  a  point   he  placed   over  every  7ith  digit  of  any  number, 

beginning  at  the  units'  place  and  exlendiyig  in  either  direction,  the 
number  of  poiyits  shows  the  number  of  digits  in  its  nth  root. 

324.  Let  a,  b,  c,  and  n  represent  positive  integers. 
Bv^285    (a  +  &  +  c)»-a-^[a  +  (6+c)]--a- 

^na^~\b  +  c)  +  ^-.^ 

That  is,  if  the  remainder  obtained  by  subtracting-  d^  from 
(aH-  6  4-  cy  be  divided  by  7ia^~^,  the  quotient  is  greater  than  b. 

In  like  manner,  if  the  remainder  obtained  by  subtracting  a** 
from  (a  +  by  be  divided  by  na""^,  the  quotient  is  greater  than  b. 

325.  It  is  evident,  from  §§  323  and  324,  that  the  nth  root  of 
a  positive  integral  perfect  ?ith  power  may  be  found  by  a  process 
similar  to  that  employed  in  §§  302  and  315. 

The  general  rule  will  be  as  follows : 

Point  the  number  in  accordance  with  the  rule  of  §  323,  and  let 
the  number  of  digits  in  the  root  be  m. 

Find  the  greatest  perfect  nth  power  in  the  left-hayid  period,  and 
write  its  nth  root  as  the  first  digit  of  the  root. 

Raise  the  part  of  the  root  already  found,  with  m  —  1  ciphers 
annexed,  to  the  nth  power,  and  subtract  the  result  from  the  given 
number. 

Raise  the  part  of  the  root  already  found,  with  m  —  1  ciphers 
annexed,  to  the  (n  —  l)th  power,  and  multiply  the  result  by  n. 

Divide  the  remainder  by  this  number. 

If  the  quotient  is  a  number  ichose  integral  part  contains  m  —  1 
digits,  write  its  first  digit  as  the  next  digit  of  the  root;  otherwise, 
write  0  a.9  the  next  root-digit. 


INVOLUTION  AND   EVOLUTION  203 

Raise  the  part  of  the  root  already  found,  with  m  —  2  ciphers 
annexed,  to  the  nth  power,  and  subtract  the  result  from  the  given 
number. 

The  above  process  is  to  be  repeated  until  there  is  no 
remainder;  the  only  change  being  that,  in  the  successive 
applications  of  the  rule,  m  —  2,  m  —  3,  etc.,  are  written  in  place 
of  m  —  1  in  the  fourth,  sixth,  and  seventh  paragraphs. 

The  rule  may  be  used  to  find  the  wth  root  of  any  number. 

Ex.  Find  the  cube  root  of  34550.415593. 

In  this  case,  n  =  3  and  m  =  4. 

34550. 415593  |  32.57 
3000^  =  27000000000 
3  X  30002  =  27000000)7550415593(200+ 

3200^  =  32768000000 
3  X  3200^  =  30720000)1782415593(50+ 

3250^  =  34328125000 
3x3250^=   31687500)222290593(7+ 
3257^  =  34550415593 

Hence,  the  required  root  is  32.57. 

Some  of  the  ciphers  may  be  omitted  in  practice. 

It  sometimes  happens  that,  on  raising  the  part  of  the  root  already  found 
to  the  nth.  power,  the  result  is  greater  than  the  given  number ;  in  such  a 
case,  the  digit  of  the  root  last  obtained  is  too  great,  an(J  one  less  must  be 
substituted  for  it. 

326.  Let  m  and  n  be  positive  integers,  and  a  a  perfect  power 
of  the  degree  mn. 

By  §157,  C^~ar-  =  a,  (1) 

and  (VV«)-=Va.  (2) 

Raising  both  members  of  (2)  to  the  nth.  power, 

{^WaY^  =  a.  (3) 

From  (1)  and  (3),  by  §  163, 

mn  /—         ^/  n/— 

"va  =  ^  va; 
for  each  of  these  expressions  is  the  mnih.  root  of  a. 


204  ADVANCED  COURSE  IN  ALGEBRA 

That  is,  the  mnth  root  of  a  (§  162)  equals  the  mth  root  of  the 
nth  root  of  a. 

The  above  is  only  true  of  principal  roots. 

It  follows  from  the  above  that  the  fourth  root  of  a  perfect 
power  of  the  fourth  degree  equals  the  square  root  of  the  square 
root  of  the  expression. 

The  sixth  root  of  a  perfect  power  of  the  sixth  degree  equals 
the  cube  root  of  the  square  root  of  the  expression ;  etc. 

In  like  manner,  if  m,  n,  and  p  are  positive  integers,  and  a 
a  perfect  power  of  the  degree  mnp, 

"Ta=V(VvS); 
and  so  on. 

327.  Let  m,  n,  and  r  be  positive  integers,  and  a  a  rational 
number  whose  mth  power  is  positive  if  n  is  even. 


By  §  157,  {-^/a^y  =  a-. 

Raising  both  members  to  the  rth  power,  we  have 

{■y/'ory  =  a'"'".  (1) 

Also,  CV^O"  =  »""■•  (2) 

From  (1)  and  (2),       ( Vo^)"'*  =  (Va^)"'. 
Taking  the  nrth  root  of  both  members  (§  163), 

=  Va    ,  and  Va     =  va  .  « 


This  means  that  the  principal  ?ith  root  of  a"*  is  equal  to  the  principal 
nrth  root  of  a"**". 

(The  general  theorems  of  evolution,  in  §§  163,  164,  165,  292,  293,  326, 
and  327,  were  there  proved  only  ioT  principal  roots. 

That  of  §  326  is  only  true  for  such  roots. 

The  others  are  true  for  certain  values  of  the  roots  which  are  not  prin- 
cipal roots  ;  take,  for  example,  the  equation 

Vab=^Va^y/h  {%  165). 

If  w  =  2,  «  =  4,  &  =  9,  it  becomes  \/4  x  9  =  Vi  x  \/9.  

The  last  equation  is  true  when  the  value  +  6  is  taken  for  V4  x  9, 


for  Vi,  and  —  3  for  V9  ;  also,  when  the  value  —  6  is  taken  for  y/4t  x  9, 
+  2  for  \/4,  and  -  3  for  V9.) 


INVOLUTION   AND  EVOLUTION  205 

EXERCISE  41 
Find  the  values  of  the  following : 
1.   v'(125  a3)2.  2.   v^Cie  xY^)^-  3.   v^(- 243  a5&25cio)8. 

Find  by  inspection  the  values  of  the  following  : 


343  6«  >  812/8  >'32p30 

Find  the  square  roots  of  : 

7.  49  a*  +  16  64  +  14  a8&  -  8  aS^  _  55  ^252. 

8.  16  x^  +  9  y"^  +  25  z^  -  24:  xy- 40  XZ  + 30  yz. 

9  2/8     8  2/6     60?/*     5y2"*"25' 

10.  4  x2  -  31  a;4  +  4  -  30  a;5  +  44  x^  +  25  x^  -  16  x. 

11.  99729.64.  12.    64.91041489.  13.    .0063138916. 

Find  the  cube  roots  of : 

14.   27  x^  -  27  x^y  -  99  ic*?/^  +  71  x^y^  +  132  x'^y^  -  48  xy^  -  64  y^. 

15     8a«     12  g^     10  a      5,56       6^         53 

16.  12  x*  -  6  x8  -  27  X  +  a:9  +  62  x3  +  9x7  +  27  -  45  x^  +  13  x^  -  45  x^. 

17.  201.230056.  18.    8831234.763.  19.    .537764475968. 

20.  Find  the  fourth  root  of 

16  x8  +  32  x'  -  72  x6  -  136  x^  +  145  x*  +  204  x^  -  162  x2  -  108  x  +  81. 

21.  Find  the  sixth  root  of 

1  +  12  X  +  54  x2  +  100  x3  +  15  x4  -  168  x^  -  76  x^  +  168  x' 
+  15  x8  -  100  x9  +  54  xw  -  12  x"  +  xi2. 

22.  Find  the  fifth  root  of 

32  xi'>  -  80  x9  +  240  x^  -  360  x?  +  570  x^  -  561  x^  +  570  x* 
-  360  x3  +  240  a;2  _  80  X  +  32. 

23.  Find  the  fourth  root  of  888.73149456. 

24.  Find  the  sixth  root  of  .009229812275335744. 

25.  Find  the  fifth  root  of  8472886.09443. 


206       ADVANCED  COURSE  IN  ALGEBRA 


XVI.    INEQUALITIES 

328.  An  Inequality  is  a  statement  that  one  of  two  expres- 
sions is  greater  or  less  than  another. 

The  First  Member  of  an  inequality  is  the  expression  to  the 
left  of  the  sign  of  inequality,  and  the  Second  Member  is  the 
expression  to  the  right  of  that  sign. 

Any  term  of  either  member  of  an  inequality  is  called  a  term 
of  the  inequality. 

Two  or  more  inequalities  are  said  to  subsist  in  the  same  sense 
when  the  first  member  is  the  greater  or  the  less  in  both. 

Thus,  a  >  6  and  c>  d  subsist  in  the  same  sense. 

PROPERTIES   OF  INEQUALITIES 

329.  An  inequality  will  continue  in  the  same  sense  after  the 
same  number  has  been  added  to,  or  subtracted  from,  both 
members. 

This  follows  from  §  23,  which  is  supposed  to  hold  for  all 
values  of  the  letters  involved. 

330.  It  follows  from  §  329,  that  a  term  may  be  transposed 
from  07ie  member  of  an  inequality  to  the  other  by  changing  its 
sign. 

If  the  same  terra  appears  in  both  members  of  an  inequality  affected 
with  the  same  sign,  it  may  be  cancelled. 

331.  If  the  signs  of  all  the  terms  of  an  inequality  be  changed, 
the  sign  of  inequality  must  be  reversed. 

For  consider  the  inequality  a  —  b>c  —  d. 

Transposing  every  term,        d  —  c>b  —  a.  (§  330) 

That  is,  b  —  a<d  —  c. 

332.  Let  a  —  6  be  a  positive  number. 
Then,   (a  —  b)-{-b>b;  that  is,  a > b. 
Again,  let  a  —  6  be  a  negative  number. 
Then,   (a  —  b)-{-b<b;  that  is,  a  <  6. 


INEQUALITIES  207 

333.  An  inequality  will  continue  in  the  same  sense  after  both 
members  have  been  multiplied  or  divided  by  the  same  positive 
number. 

For  consider  the  inequality  a>b. 

By  §  332,  a  —  6  is  a  positive  number. 

Hence,  if  m  is  a  positive  number,  each  of  the  numbers 

m(a  —  b)  and  , 

or,  ma  —  mb  and ,  is  positive. 

mm 

Therefore,  ma  >m?>,  and  —  >  — 

m     m 

334.  It  follows  from  §§  331  and  333  that  if  both  members  of 
an  inequality  be  multiplied  or  divided  by  the  same  negative  num- 
ber, the  sign  of  inequality  must  be  reversed. 

335.  If  any  number  of  inequalities,  subsisthig  in  the  same 
sense,  be  added  member  to  member,  the  resulting  inequality  will 
also  subsist  in  the  same  sense. 

For  consider  the  inequalities  a>b,  a'  >b',  a">b",  •••. 

Then  each  of  the  numbers,  a  —  b,  a' —  b',  a"  —  b",  •••,  is 
positive. 

Therefore,  their  sum 

a-b-{-a'-b'  -ha"  -b" +*••', 

or,  a  +  a'  +  a"  4- (&  +  &'  +  b"  +  .••), 

is  a  positive  number. 

Whence,      a-fa'  +  a"  +  •••  >  b  ■{- b'  -\- b"  -{-  •••. 

336.  If  two  inequalities,  subsisting  in  the  same  sense,  be 
subtracted  member  from  member,  the  resulting  inequality  does 
not  necessarily  subsist  in  the  same  sense. 

Thus,  it  a>b  and  a'  >b',  the  numbers  a  —  b  and  a'  —  &'  are 
positive. 

But  (a  —  b)  —  (a'  —  b'),  or  its  equal  (a  —  a')  ~  (b  —  b'),  may 
be  positive,  negative,  or  zero ;  and  hence  a  —a'  may  be  greater 
than,  less  than,  or  equal  to  6  —  b'. 


208  ADVANCED  COURSE  IN  ALGEBRA 

337.  If  a  >  &  and  a'  >b',  and  each  of  the  numbers  a,  a', 
b,  b',  is  positive,  then  ^^t      ^^, 

For  since  a—b,  a'  —  b',  a,  and  b'  are  positive  numbers,  each 
of  the  numbers        ^^^,  _  ^,^  ^^^  b'(a-b) 
is  positive. 

Then,  aa'>ab',  and  ab'>bb'. 

Then  by  §  335,       aa'  +  ab'  >  ab'  +  bb'. 
Whence,  aa'  >  bb'. 

338.  If  we  have  any  number  of  inequalities  subsisting  in 
the  same  sense,  as  a  >  6,  a'  >b',  a"  >b",  •••,  and  each  of  the 
numbers  a,  a',  a",  •••,  b,  b',  b",  •••,  is  positive,  then 

aa'a""'>bb'b"'-  - 
For  by  §337,  aa'>bb'. 

Also,  a"  >  &". 

Whence,  aa'a"  >  &6'6"  (§  337). 

Continuing  the  process  with  the  remaining  inequalities,  we 
obtain  finally  aa'a"  ...  >  66'6"  ...  • 

339.  It  follows  from  §  338  that,  if  a  is  >  b,  and  a  and  b  are 
positive  numbers,  and  n  a  positive  integer,  then 

a"  >  b\ 

340.  If  71  is  a  positive  integer,  and  a  and  b  perfect  nth 
powers  such  that  a  is  >  6,  then 

■Va>^b. 

For,  if  "Va  were  <  -\/b,  raising  both  members  to  the  nth 
power,  we  should  have  a  <  b.  (§  339) 

And,  if  -\/a  =  -\/by  a  would  be  equal  to  b. 

Both  of  these  conclusions  are  contrary  to  the  hypothesis 
that  a  is  >  6. 

Hence,  -\/«  >  'Vb, 


INEQUALITIES  ^09 

341.   Examples. 

1.  Find  tlie  limit  of  x  in  the  inequality 

Multiplying  both  members  by  3  (§  333),  we  have 

21a;  -  23  <  2a; +  15. 
Transposing  (§  330),  and  uniting  terms, 

19x<38. 
Dividing  both  members  by  19  (§  333), 

a;<2. 

2.  Find  the  limits  of  x  and  y  in  the  following : 


3a;  +  22/>37. 

(1) 

2a; +  32/ =  33. 

(2) 

9a;  +  6?/ >  111. 

4a; +  6  2/=   ^^• 

5x>    45,  and  a;  >  9. 

6  a;  +  4  2/  >  74. 

6  a;  +  9  2/  =  99. 

Multiply  (1)  by  3, 

Multiply  (2)  by  2, 

Subtracting  (§  329), 

Multiply  (1)  by  2, 

Multiply  (2)  by  3, 

Subtracting,  —5y>  —25. 

Divide  both  members  by  —  5,      2/  <  5  (§  334). 

3.  Between  what  limiting  values  of  a;  is  a;^  —  4  a;  <  21  ? 

a;2-4a;is<21  if  a;2-4a;-21  is<0. 
That  is,  if  (x  +  3)(a;  —  7)  is  negative. 
Now  (x  +  3)(a;  —  7)  is  negative  if  x  is  between  —  3  and  7. 
Hence,  a;^  —  4  a;  is  <  21  if  oj  is  >  —  3,  and  <  7. 

4.  Prove  that  if  a  and  b  are  positive  numbers, 

b     a 

Wehave(a-6)2<0;  ot,  a' -2  ab-^b^-^O. 
Transposing  —2ab,  a^  +  6^  <  2  ab. 
Dividing  each  term  by  ab, 

b     a 


210  ADVANCED   COURSE   IN.  ALGEBRA 

5.  Prove  that,  if  a  and  h  are  unequal  positive  numbers, 

Wehave  (a-6)2>0;  or,  a^ -2  a& -f-&'>0. 
Transposing  —  ah,  o?  —  ah  +  6'^  >  ah. 
Multiplying  both  members  by  the  positive  number  a  +  6, 
a^^h^>a%^-h''a. 

6.  Prove  that,  if  a,h,  and  c  are  unequal  positive  numbers, 

2(a3  +  63  _^  c^)  >  a^?)  +  h'^a  +  6'c  +  c^h  +  c^a  +  a^c. 
By  Ex.  5,  d'  +  h^>a^h  +  h''a, 

¥  +  c'>h'c+c% 
and  c^  -|~  a^  >  c^a  +  a^c. 

Adding,  2(a3  +  6^  +  c=^)  >  a'b  +  ft^a  +  b'c  +  c^ft  4-  c^a  +  a'c. 

EXERCISE  42 

Fiud  the  limits  of  x  in  the  following : 

1.  (2x-3)3_71>4x(2x-5)(a;-2). 

2.  (2  -  3x)(3  -  a:)+  4x  +  39  >3  +(2  +  3a;)(x  +  3). 

3.  (x  -  l)(x  -  2)  (x-  3)<(a;  -  5)(x  +  6)(x  -  7). 

4.  a^(x- l)<2  62(2x- l)-a&,  if  a-26  ispositive. 
^    X  -  m  ,  2  ^  x_+n^    •£  ^  ^j^(j  ^  ^j.g  positive  and  w  <  n. 


n  m 

Find  the  limits  of  x  and  y  in  the  following : 
5x  +  6  2/<45. 


^'     '  3x-42/  =  -ll. 


J  7x-4?/>41. 


8.  Find  the  limits  of  x  when 

3  X  -  11  <  24  -  11 X,  and  5  x  +  23  <  20  x  +  3. 

9.  If  6  times  a  certain  positive  integer,  plus  14,  is  greater  than  13 
times  the  integer,  minus  63,  and  17  times  the  integer,  minus  23,  is  greater 
than  8  times  the  integer,  plus  31,  what  is  the  integer  ? 

10.  If  7  times  the  number  of  houses  in  a  certain  village,  plus  33,  is 
less  than  12  times  the  number,  minus  82,  and  9  times  the  number,  minus 
43,  is  less  than  5  times  the  number,  plus  61,  how  many  houses  are  there  ? 


INEQUALITIES  211 

11.  A  farmer  has  a  number  of  cows  such  that  10  times  their  number, 
plus  3,  is  less  than  4  times  the  number,  plus  79  ;  and  14  times  their  num- 
ber, minus  97,  is  greater  than  6  times  the  number,  minus  5.  How  many 
cows  has  he  ? 

12.  Between  what  limiting  values  of  ic  is  a;^  4.  3  x  <  4  ? 

13.  Between  what  limiting  values  of  x  is  2  aj^  +  1.3  x  >  24  ? 

14.  Between  what  limiting  values  of  ic  is  6  x^  <  19  x  —  10  ? 


Prove  that,  for  any  values  of  x, 


49 


15.    9x2  +  25<30jc.  16.    x(x-3)<| 

2i      lb 

Prove  that,  for  any  values  of  a  and  6, 

17.  (4a  +  36)(4a-36)<66(4a-3&). 

18.  a4  4.  ^4^  2  ah{d?-  -  ah  ■\-  V^\ 

Prove  that,  if  all  the  letters  represent  unequal  positive  numbers, 

19.  d^  +  a2&  +  a52  +  53  >  2  a&(a  +  &). 

20.  a2  +  62  _f.  c2  >  «&  +  6c  +  ca. 

21.  aW  +  &2c2  -I-  c2a2  >  d?'hc  +  V^ca  +  <^Mh. 

22.  (a  +  6  -  c)2  +  (&  +  c  -  a)2  +  (c  +  a  -  hy  >  a&  +  6c  +  ca. 

23.  a26  +  a62  +  62c  +  6c2  +  c2a  +  ca2  >  6  a 6c. 

24.  (a2  +  62  +  C2)  (X2  +  ^2  4.  ^2)^  (Qja;  +  6?/  +  czf. 


212  ADVANCED   COURSE  IN  ALGEBRA 


XVII.  SURDS.  THEORY  OF  EXPONENTS 

342.  Meaning  of  V2. 

It  is  impossible  to  find  a  rational  number  (§  51)  whose 
square  shall  equal  2 ;  but  we  can  find  two  rational  numbers, 
which  shall  differ  from  each  other  by  less  than  any  assigned 
number,  however  small,  whose  squares  shall  be  less,  and  greater 
than  2,  respectively. 

For,  writing  the  squares  of  the  consecutive  integers  1,  2,  etc., 
we  have  1^  =1,  2^  =  4,  etc. 

Hence,  1  and  2  are  two  numbers  which  differ  by  1,  and 
whose  squares  are  less  and  greater  than  2,  respectively. 

Again,  1.12  =  1.21,  1.22=1.44,  1.32=1.69,1.42  =  1.96,  1.5'  = 
2.25,  etc. 

Hence,  1.4  and  1.5  are  two  numbers  which  differ  by  .1,  and 
whose  squares  are  less  and  greater  than  2,  respectively. 

Again,  1.412=1.9881,  1.422  =  2.0164,  etc. 

Hence,  1.41  and  1.42  are  two  numbers  which  differ  by  .01, 
and  whose  squares  are  less  and  greater  than  2,  respectively. 

By  suificiently  continuing  the  above  process,  we  can  find  two 
numbers  which  shall  differ  from  each  other  by  less  than  any 
assigned  number,  however  small,  whose  squares  shall  be  less 
and  greater  than  2,  respectively. 

343.  The  successive  numbers,,  in  the  illustration  of  §  342, 
whose  squares  are  less  than  2,  are  1,  1.4,  1.41,  etc. ;  and  the 
numbers  whose  squares  are  greater  than  2,  are  2,  1.5,  1.42,  etc. 

If  each  series  be  continued  to  r  terms,  the  difference  between 
the  rth  terms  of  the  two  series  is 

1 

which  can  be  made  less  than  any  assigned  number,  however 
small,  by  sufficiently  increasing  r. 

Therefore,  the  rth  terms  of  the  two  series  approach  the  same 
limit  (§  245),  when  r  is  indefinitely  increased. 


SURDS.    THEORY  OF  EXPONENTS  213 

This  limit  is  taken  as  the  definition  of  V2. 

344.  In  general,  if  n  is  any  positive  integer,  and  a  a  rational 
number  (§  51),  which  is  not  a  perfect  power  of  the  nth.  degree, 
and  which  is  positive  if  n  is  even,  it  is  impossible  to  find  a 
number  whose  nth  power  shall  equal  a. 

We  can  find,  however,  two  rational  numbers  which  shall  dif- 
fer from  each  other  by  less  than  any  assigned  number,  how- 
ever small,  whose  nth  powers  shall  be  less  and  greater  than  a, 
respectively. 

345.  If  n  and  a  have  the  same  meaning  as  in  §  344,  and  a^ 
tta,  a^,  etc.,  is  a  series  of  rational  numbers  whose  nth  powers 
are  less  than  a,  and  a'l,  a\,  a'g,  etc.,  a  series  of  rational  num- 
bers whose  nth  powers  are  greater  than  a,  such  that  a\  ~  ai  = 
1,  a'2  ~  a^  =  .1,  a'g  ~  cig  =  .01,  etc.,  we  may  show,  as  in  §  343, 
that  the  rth  terms  of  the  two  series  approach  the  same  limit, 
when  r  is  indefinitely  increased. 

This  limit  is  taken  as  the  definition  of  Va. 
The  expression  Va  is  called  a  Surd. 

The  symbol  ~  signifies  the  difference  of  the  numbers  between  which 
it  is  placed. 

346.  In  the  illustration  of  §  342,  we  also  have  (  - 1)^  =  1, 
(-2)2  =  4,   etc. 

It  is  therefore  possible  to  find  two  negative  rational  numbers, 
which  shall  differ  from  each  other  by  less  than  any  assigned 
number,  however  small,  whose  squares  shall  be  less  and  greater 
than  2,  respectively. 

This  is  also  the  case  with  every  surd  of  the  form  Va,  when 
n  is  even. 

347.  If  a  is  positive,  and  <Xi,  ag?  etc.,  and  a\,  a\,  etc.,  be  taken 
with  positive  signs,  we  shall  call  the  limit  approached  by  the 
rth  terms  of  the  two  series  of  §  345,  when  r  is  indefinitely 
increased,  the  principal  nth  root  of  a. 

If  a  is  negative,  we  shall  call  the  negative  limit  approached 
by  the  rth  terms  of  the  series,  when  r  is  indefinitely  increased, 
the  principal  nth  root  of  a. 


214  advancp:d  course  in  algebra 

operations  involving  surds 

348.  It  is  necessary  to  define  Addition  and  Multiplication, 
when  any  or  all  of  the  numbers  involved  are  surds. 

Let  n  and  p  be  positive  integers. 

Let  a  be  a  rational  number  which  is  not  a  perfect  power  of 
the  nth.  degree,  and  is  positive  if  n  is  even. 

Let  &  be  a  rational  number  which  is  not  a  perfect  power  of 
the  pth  degree,  and  is  positive  if  p  is  even. 

Let  c/j,  ttg,  •••,  a^,  '-',  be  a  series  of  rational  numbers  whose 
nth  powers  are  less  than  a,  and  a\,  a\,  •••,  a'^,  •••,  a  series 
whose  nth  powers  are  greater  than  a,  such  that 

Let  hi,  &2J  •••?  ^r)  •••?  be  a  series  of  rational  numbers  whose 
pth  powers  are  less  than  h,  and  h\,  h\,  •••,  6^?  •••?  a  series 
whose  pth  powers  are  greater  than  6,  such  that 

h\^h,    =1,    h',^h,--=l,    ...,    6V~6,=:(.1X-S    .... 

Then  to  add  -y/b  to  Va,  is  to  find  the  limit,  when  r  is  indefi- 
nitely increased,  of  a^  +  b^. 

To  multiply  Va  by  V^  is  to  find  the  limit,  when  r  is  indefi- 
nitely increased,  of  a^  x  b^. 

349.  A  meaning  similar  to  the  above  is  attached  to  any 
expression,  which  is  not  a  rational  number,  and  which  is  the 
result  of  any  finite  number  of  the  following  operations  per- 
formed upon  one  or  more  rational  numbers,  provided  that,  in 
any  indicated  root,  the  number  under  the  radical  sign  is  posi- 
tive if  the  index  of  the  root  is  even : 

Addition;  Subtraction;  Multiplication ;  Division ;  raising  to 
any  positive  integral  power ;    extracting  any  root. 

350.  We  will  now  show  how  to  prove  the  laws  of  §§  12  and 
14,  when  any  or  all  of  the  letters  involved  represent  surds. 

Let  it  be  required,  for  example,  to  prove  the  Commutative 
Law  for  Multiplication  (§  14)  with  respect  to  the  product 
of  two  surds,  "v/a  and  {/&,  where  n,  p,  a,  and  b  have  the  same 
meanings  as  in  §  348. 


SURDS.     THEORY   OF   EXPONENTS  215 

Or,  to  prove  \^a  x  ^b=-\/b  x  -Va. 

With  the  notation  of  §  348,  -Va  x  -\/b  is  the  limit,  when  r  is 
indefinitely  increased,  of  a^  x  b^. 

Also,  -\/&  X  -\/cL  is  the  limit,  when  r  is  indefinitely  increased, 
of  b^  X  a,.. 

By  §  14,  since  a^  and  br  are  rational  numbers, 
a^xK  =  brX  a^. 

Then,  a^  x  b^  and  &^  X  a^  are  functions  of  r  which  are  equal 
for  every  positive  integral  value  of  r;  and,  by  §  252,  their 
limits  when  r  is  indefinitely  increased  are  equal. 

Hence,  -y/a  X  V&  =  V&  X  Va. 

This  simply  means  that  the  product  of  the  principal  nth  root  of  a 
(§  347)  by  the  principal  pth  root  of  b  is  equal  to  the  product  of  the  princi- 
pal pth  root  of  b  by  the  principal  nth  root  of  a.     (Compare  §  162.) 

A  similar  interpretation  must  be  given  to  every  result  involving  surds. 

In  like  manner,  the  remaining  laws  of  §§12  and  14  may  be 
proved  to  hold,  when  any  or  all  of  the  letters  involved  repre- 
sent surds. 

351.  Since  the  remaining  results  of  Chap.  I,  and  the  results 
in  Chaps.  II  to  XVI,  inclusive,  are  formal  consequences  of  the 
laws  of  §§12  and  14,  it  follows  that  every  statement  or  rule, 
in  these  chapters,  in  regard  to  expressions  where  any  letter 
involved  represents  any  rational  number,  holds  equally  when 
the  letter  represents  a  surd. 

This  is  also  the  case  when  the  letter  represents  any  number  of  the 
form  described  in  §  349. 

APPROXIMATE  ROOTS 

352.  Any  one  of  the  successive  numbers,  in  the  example  of 
§  342,  is  called  an  approximate  square  root  of  2. 

In  general,  any  one  of  the  numbers  %,  a2,  etc.,  or  a'l,  a'g,  etc., 
in  §  345,  is  called  an  approximate  nth  root  of  a. 

353.  The  successive  numbers,  in  the  example  of  §  342,  whose 
squares  are  less  than  2,  may  be  obtained  by  regarding  2  as  a 
perfect  square,  and  applying  the  rule  of  §  304. 


216  ADVANCED  COURSE   IN  ALGEBRA 

'  2.060600  I  1.414 

1 


24 

100 
96 

281 

400 

281 

2824 

11900 
11296 

604 

The  process  may  be  continued  to  any  desired  extent. 

In  like  manner  the  rule  of  §  317  may  be  used  to  find  an 
approximate  cube  root  of  a  number  (Note,  §  312)  which  is  not 
a  perfect  cube ;  and  the  rule  of  §  325  may  be  used  to  find  an 
approximate  nth  root  of  a  number  (Note,  §  323)  which  is  not  a 
perfect  power  of  the  nth  degree. 

The  considerations  in  §§  306  and  320  apply  equally  to  ap- 
proximate square  and  cube  roots. 

354.  To  find  an  approximate  root  of  a  fraction  whose  terms 
are  positive  integers  expressed  in  Arabic  numerals,  whose 
denominator  is,  and  whose  numerator  is  not,  a  perfect  power 
of  the  degree  denoted  by  the  index  of  the  required  root,  we 
may  divide  the  required  approximate  root  of  the  numerator  by 
the  required  root  of  the  denominator  (§  293). 

If  the  denominator  is  not  a  perfect  power  of  the  degree 
denoted  by  the  index  of  the  required  root,  the  fraction  should 
be  reduced  to  an  equivalent  fraction  whose  denominator  is  a 
perfect  power  of  this  degree. 

Thus,  let  it  be  required  to  find  the  square  root  of  f  approxi- 
mately, to  four  decimal  places. 

V|  =  V5=f(§293)=2:M9i  =  .6m.... 

EXERCISE  43 

Find  the  approximate  value  of  each  of  the  following  to  five  decimal 
places : 

1.    \/5.  2.    VTS.  8.    V:85l.  4.    \/:003. 


SURDS. 

THEORY 

OF   EXPONENTS 

217 

5. 

vi- 

8. 

4 

11.    </J. 

14. 

f 

6. 

4 

9. 

^4. 

12.    y/M. 

15. 

f 

7. 

>lk 

10. 

^IT. 

-f 

16. 

€• 

THE  THEORY  OF  EXPONENTS 

355.  In  the  preceding  portions  of  the  work,  an  exponent 
has  been  considered  only  as  a  positive  integer. 

Thus,  if  m  is  a  positive  integer, 

a"^  =  a  X  a  X  a  X  "•  to  m  factors.  (§  60) 

356.  We  have  proved  the  following  results  to  hold  when  m 
and  n  represent  positive  integers,  and  a  any  rational  number : 

a'^Xa''  =  a'^+'*  (§  85).  (1) 

(a"^)"  =  a*""  (§  128).  (2) 

357.  It  is  necessary  to  employ  exponents  which  are  not  posi- 
tive integers ;  and  we  now  proceed  to  define  those  forms  of  ex- 
ponent which  are  rational  numbers,  but  not  positive  integers. 

In  determining  what  meanings  to  assign  to  the  new  forms, 
it  will  be  convenient  to  have  them  such  that  the  above  law  for 
multiplication  shall  hold  with  respect  to  them. 

We  shall  therefore  assume  equation  (1),  §  356,  to  hold,  what- 
ever number  is  represented  by  a,  for  all  rational  values  of  m 
and  n,  including  the  case  where  either  m  or  n  is  zero';  and  find 
what  meanings  must  be  attached  in  consequence  to  fractional, 
negative,  and  zero  exponents. 

358.  Meaning  of  a  Fractional  Exponent. 

p 
Required  the  meaning  of  a',  where  p  and  q  represent  positive 

integers. 

If  (1),  §  356,  is  to  hold  for  all  rational  values  of  m  and  n, 

we  have 

-  ?  „  -+-+•••  to  ?  terms  -xq 

a'  X  a'  X  •  •  •  to  g  factors  =  a«  *  =  a'     =  a^. 


218  ADVANCED  COURSE   IN   ALGEBRA 


p 


That  is,  (a«)'  =  a^. 

p         ■ 

Whence,  by  §  157,  a^=  ^/a^. 

p 
We  shall  then  define  a^  as  being  the  gth  root  of  a^. 

For  example,  a^  =  -\/a'^j  a^  =  Va^;  a^  =  Va;  etc. 

p 
We  shall  throughout  the  remainder  of  the  work  regard  a^  as  being  the 

principal  gth  root  of  a^. 

359.  Meaning  of  a  Zero  Exponent. 

Required  the  meaning  of  a°. 

By  §  356,  (1),  if  m  is  any  rational  number 

Whence,  a^  =  —  =  1. 

We  shall  then  define  a^  as  being  equal  to  1. 

360.  Meaning  of  a  Negative  Exponent. 

Required  the  meaning  of  a~%  where  s  represents  a  positive 
integer  or  a  positive  fraction. 

By  §  356,  (1),  a-"  x  a"  =  a-'+'  =  a«  =  1  (§  359). 

Whence,  a~' =  — 

a' 

We  shall  then  define  a~^  as  being  equal  to  1  divided  by  a*. 
For  example,  a~'^  =  —  ;  a"^  =  —  ;  3  x~^y~^  =  — -  ;  etc. 

361.  It  follows  from  §  360  that 

Any  factor  of  the  numerator  of  a  fraction  may  he  transferred 
to  the  denominator,  or   ayiy  factor   of  the   denominator  to   the 
^  numerator,  if  the  sign  of  its  exponent  he  changed. 


Thus,  —   may  be  written  in  the  forms 
cd^ 

W        d'hh-'     aH-^     . 
-,  etc. 


a-'cd''       d'    '     h-'c- 


SURDS.    THEORY  OF   EXPONENTS  219 

362.  We  will  now  prove  that,  with  the  definitions  of  §§  358 
and  360,  equation  (1),  §  356,  holds  for  all  rational  valnes  of  m, 
n,  and  a,  provided  that  a""  and  a"  are  rational  numbers  or  surds. 

It  will  be  understood  that,  in  all  fractional  exponents,  the 
results  are  limited  to  principal  roots. 

•  I.  Let  m  and  n  be  fractions  of  the  form  ^  and  -,  respectively, 
where  p,  q,  r,  and  s  represent  positive  integers. 

By  §  358, 


a^  Xa'  =  V~a^  X  Va'" 

=  V^xV^ 

(§  327) 

=  Va^'xa«'- 

(§  165) 

=  Va^''+^'- 

(§85) 

ps+qr              p      T 

=  a  ''    =a'   '. 

We  have  now  proved  that  (1),  §  356,  holds  when  m  and  n 
represent  any  positive  rational  numbers. 

II.  Let  m  be  rational  and  positive,  and  let  n  =  —  g,  where  q 
is  rational,  positive,  and  less  than  m. 

By  §§  85,  or  362,  I,  a"'-^  x  a'  =  a"'-«+^  =  oT. 

Whence,  a"^-^  =  ^  =  a"*  X  cr^  (§  360). 

a* 

That  is,  a"*  X  a'^  =  oT''^. 

In  like  manner,  the  law  may  be  proved  to  hold  when  n  is 
rational  and  positive,  and  m=^—p,  where p  is  rational,  positive, 
and  less  than  n. 

III.  Let  m  be  rational  and  positive,  and  let  n  =  q,  where  q 
is  rational,  positive,  and  greater  than  m. 

By  §  361,  a-  x  a''  =  -J—  =  -i—  (§  362,  II)  =  a«-'. 

In  like  manner,  the  law  may  be  proved  to  hold  when  n  is 
rational  and  positive,  and  m  =  —  p,  where  p  is  rational,  positive, 
and  greater  than  n. 


220  ADVANCED  COURSE  IN  ALGEBRA 

ly.   Let  m  —  —p,  and  n=  —  q,  where  p  and  q  are  rational 
and  positive. 

Then,  a'P  x  a"'  =  —  =  —  (§§  85,  or  362,  I)  =  a"^"*. 

Hence,  a™  x  f^  =  »*"+"  for  all  rational  values  of  m  and  n. 
For  example,  a^  x  a~^  —  a}~^  =  a~^; 

axa^  =  o}^^  =  a2 ;  etc. 

363.   We  have  for  all  rational  values  of  a,  m,  and  n,  provided 
that  a"*  and  a"  are  rational  numbers  or  surds, 

^m-n  y^gn^  ^m-n+n  (§§   35^  3(32)  =,  ^« 

Whence,  —  =  a"*-^ 


a" 

_3 

a 


Por  example,  ^  =  a  ^  ^  =  a  ^; 

^  =  a^*'  =  ai,  etc. 


a  - 


364.  We  will  now  prove  that  equation  (2),  §  356,  holds  for 
all  rational  values  of  a,  m,  and  n,  provided  that  a"*  and  a*""  are 
rational  numbers  or  surds. 

In  all  fractional  exponents,  the  results  are  limited  to  princi- 
pal roots. 

I.  Let  91  be  a  positive  integer. 

Then,     {ary  =  a"*  x  a*"  X  a"*  X  •••  to  ti  factors 

^  ^m+m+m+...  ton  terms   (§§    §5^  ^Q^)  =  0^"". 

II.  Let       n  =^,  where  p  and  q  are  positive  integers. 

p  mp 

Then,     (a'")'  =  i/Jary  =  i/aF^  (§  364,  I)  =  a^ 
HI.   Let     n  =  —  s,  where  s  is  rational  and  positive. 

Then,   (a"*)-^  =  -J—  =  —  (§  364,  I  or  II)  =  a"*"'. 

Hence,  (a*")"  =  a'""  for  all  rational  values  of  m  and  n. 
For  example,  (a^)^     =  a^^t    =  a^; 

(a^)-i    =a^x-i    =a-'*;  etc. 


SURDS.    THEORY  OF  EXPONENTS  221 

365.  To  prove  (abc  ...)"  =  a^'b^'c''  •••  for  all  rational  values  of 
a,  b,  c,  •••,  and  n,  provided  that  a%  6",  c",  •••,  are  rational  7iumbers 
or  surds. 

The  theorem  was  proved  in  §  129  for  any  positive  integral  value  of  n, 
and  in  §  165  for  any  value  of  n  of  the  form  — ,  where  m  is  a  positive 
integer.  '"^ 

I.  Let  n=^,  where  p  and  q  are  positive  integers. 

By  §  364,   l(abc  ••.)'>  =  (abc  ...)^  =  a^b^c^  ...  (§  129).         (1) 

p  p  p  p      p      p 

By  §  129,     (a'6^c«  ...)«  =  (a')«(6«)«(c^)«  •  •  •  =  a^b^c^  *  • ..  (2) 

From  (1)  and  (2), 

p  p  pp 

\_{abc  ' . -yy  =  {a^¥c^ . •  •)«'. 

p       p  pp 
Whence,         (abc  ...)*  =  a^¥c^  • . .  (§  163). 

This  means  that  the  principal  gth  root  of  {abc  .••)p  is  equal  to  the 
product  of  the  principal  gth  roots  of  a^,  6^,  c^,  •••. 

II.  Let  n  =  —  s,  where  s  is  rational  and  positive. 

Then,         (a6c  ...)-'  = -—i-— 
^  ^         {abc'-y 

^        (§§129,  or  365, 1) 


a'b'(f"' 
=  a-'b-'c--". 
Hence,  {abc  ...)"  =  a^'^c'' ...  for  all  rational  values  of  m  and  w. 

366.   Examples. 

In  the  following  examples,  every  letter  is  supposed  to  represent  a 
rational  number  such  that  every  expression  of  the  form  a«  is  a  rational 
number  or  surd. 

The  value  of  a  number  affected  with  a  fractional  exponent 
may  be  found  by  first,  if  possible,  extracting  the  root  indicated 
by  the  denominator,  and  then  raising  the  result  to  the  power 
indicated  by  the  numerator. 

1.   Find  the  value  of  (-8)i 


(-8)^  =  V(-8)2=(V-8)2  (§292)  =  (-2)2  =  4. 


222 


ADVANCED  COURSE   IN   ALGEBRA 


2.   Multiply  a  +  2a^-3a3  by  2-4a~^-6a"^. 

a  +  2j  -   3a^ 
2-4 tt"^-   6a~^ 


2  a  +  4  a  3 

-4:  J 


6a^ 

8a^  +  12 
6a^_12  +  18a"^ 


2a 


20  a*  + 18  a"^. 


3.   Divide  lSxy-'--23-{-x~iy-\-6x-y 

by  3  ic^2/^^  +  ic^  —  2  ic~^2/- 


18  xy^  -  23  +  a;~^2/  +  6  x-y 
IS  xy-^  + 6  x^jr' -12 


3  aj%~^  +  a?^  —  2  a?  ^,j 


6xh 


2x~^-3x~^y 


—  6  x^y~^  — 11  +    X  ^y-{-6  x~y 

—  6flj^2/"^—   2-\-4.x~^y 


-  9-3a;~^2/4-6£c 

-  9-3a;~^?/  +  6 


ic~y 


It  is  important  to  arrange  the  dividend,  divisor,  and  each  remainder 
in  the  same  order  of  powers  of  some  common  letter. 

/  i  o, \5 

.  4.   Expand  (  — —  Vm^j  by  the  Binomial  Theorem. 
1 


c»^ 


=  (^-1)5  _^  5  (m-^)X-  m^)  + 10  (m-^)\-  m^f 
+ 10  (m-^)2(-  m^f  +  5  (m-^)(-  m^)^ 

=  m  ^'  —  5  m~^  •  m^  + 10  m~^  •  m^  — 10  m~^  .•  m* 

-{-5m  ^  -m^  —m^ 
—  ^-¥  _  5  ^-|  _|_  10  m^^^  — 10  m^  +  5  rrfi^  —  m  ^. 


SURDS.    THEORY  OF   EXPONENTS  223 


EXERCISE  44 

It  will  be  understood,  in  the  following  set  of  examples,  that  every 
letter  used  as  the  index  of  a  root  represents  a  positive  integer,  and  every 
other  letter  a  rational  number,  such  that  every  expression  of  the  form 
a"  or  \^a  is  a  rational  number  or  surd. 

Express  with  radical  signs : 

31  251  4    1  ^  I  - 

1.   a^b^.  2.   x^y^z^.  3.   6mJn^.  4.   a^x^y*". 

Express  with  fractional  exponents : 


Express  with  positive  exponents  : 


p 


9.   x'^y-^.  10.   4a  ^b^.  11.   rri^n  2.  12.    a-»«6«c  2. 

In  each  of  the  following,  transfer  all  literal  factors  from  the  denomi- 
nator to  the  numerator : 

13.    ^-  14.    -1^.  15.    -^^.  16.   ^ 


0  y  '±y  z  a^b-^c  « 

In  each  of  the  following,  transfer  all  literal  factors  from  the  numerator 
to  the  denominator : 

17.   ^.  18.    ^l!^.  19.    2  rn~^x^  20.    ^If^. 


6 


Find  the  values  of  the  following : 

21.  (aT^)5.  23.    (a;*)"^^.  25.    126^  27.    (- 1024)i 

22.  (a-3)-9.  24.    (a"^)T^.  26.    16^.  28.   729^. 


Multiply  the  following : 

29.  a:2  _  4  a^f  _  5  _|_  6  x~^  by  2  cc"^  +  r7^  -  3  x"^. 

30.  m~^  +  2  wi-in-i  +  3  m~^w-2  by  2  m~^w-i  -  4  m~^w-2  +  6  w^.  ft  ^ 

31.  3  a^&^  +  4  a6^  -  a^&  by  6  a^6~^  -  8  a"^6"^  -  2  a~i  ^>^ '    .^ 


Divide  the  following : 

32.  TO"2n  —  5  w~i  +  4  m^ji-^  by  w^w^  —  m'^n  —  2  m*^. 

33.  a;2|/"^  -  10  xxf"^  +  9  by  Aj^  ^2x\^  -  3  x%. 

34.  atfti  _  2  62  +  a-lfti  by  a^b^  -26^  +  a"^6^. 


f** 


v> 


4 


1  ', 


|.L/ 


224       ADVANCED  COURSE  IN  ALGEBRA 

(In  the  following  ten  examples,  use  the  rules  of  Chap.  VII.) 
Find  the  value  of  : 
35.    (2a^  +  3&"*)2.  36.    ^6  m-^^  -  S  mH-^y. 

37.  (4  x^y'i  +  7  z-'^)  (4  x  V^  -  7  z-^) . 

38.  {Sx^-4.y~^y.  40.    (a^b~^  -  2a^  -  a-'^bfy. 

39.  (a-263  +  2  a35-2)3.  41.    (x^  -  3  a;^  +  2  ic~^)3. 

^^    25a-6-49m^  ^3    8a;2  +  27y~^  ^^^    a^  -  b~^ 

5  a-3  -  7  m^  2x^  +By~^  a^  +  &~* 

45.  Factor  a^  -  S  6-9  by  the  rule  of  §  177. 

46.  Factor  aJ  +  ah~^  +  b~^  by  the  rule  of  §  172. 

Expand  the  following  by  the  Binomial  Theorem  : 

47.  (a^  +  3  6~^)4.  49.    (a~^\/¥  -  b'^Vcfiy. 

48.  (zh-i-\y.  BO.  p_v5!  +  .?rV.   ■ 

Find  the 
51.   6th  term  of  (y/a^  +  v^a)".  62.   7th  term  of  (x%^  -  ^ Y*. 

53.    10th  term  of  f  2  v^w —\  ^^ 

54.  Find  the  square  root  of  aJb-^  -  6  a^6"2  +  5  ft-i  +  12  a"^  +  4  a~^&. 

55.  Find  the  cube  root  of 

a;^-  _  6  X*  +  2 1  x~^  -  44  jc"'^  +  63  x"^  -  54  x"^  +  27  x~i 

Simplify  the  following : 

56.  [^(xV2)-v^(aJ~V)]^.  60.    l^C^V^^9)^^r\ 

M  ^       s.    V^ '-V/^  CI       X^"  -  1    ,    X2«  +  1 


68.    (^^7«4!V-  •-.  62.    -^il^i:^x------  +  l. 


a.2u  4.  1        jc2 


\    Q^TO-W 


Qj  i  _  5  i       a3  _  ^i 


59    x^+y^     a;  +  y,  g3    q^  +  2  &^        7  a^&^  +  6  fe^ 


SURDS.    THEORY  OF  EXPONENTS  225 

REDUCTION  OF   SURDS 

It  will  be  understood,  in  §§  367  to  398,  inclusive,  that  every  letter  used 
as  the  index  of  a  root,  represents  a  positive  integer,  and  every  other  letter 
a  rational  number  such  that  every  expression  of  the  form  Va  is  a  surd. 

367.  If  a  surd  is  in  the  form  b-Va,  where  a  and  b  are  rational 
expressions  (§  198),  b  is  called  the  coefficient,  and  n  the  index; 
and  the  surd  is  said  to  be  of  the  nth  degree. 

368.  A  quadratic  surd  is  a  surd  of  the  second  degree. 

369.  Similar  Surds  are  surds  which  do  not  differ  at  all,  or 
differ  only  in  their  coefficients ;  as  2-\/ax^  and  3  V«^. 

Dissimilar  Surds  are  surds  which  are  not  similar^ 

370.  Reduction  of  a  Surd  to  its  Simplest  Form. 

A  surd  is  said  to  be  in  its  simplest  form  when  the  expression 
under  the  radical  sign  is  rational  and  integral  (§  63),  is  not  a 
perfect  power  of  the  degree  denoted  by  any  factor  of  the  index 
of  the  surd,  and  has  no  factor  which  is  a  perfect  power  of  the 
same  degree  as  the  surd. 

371.  Case  I.  When  the  expression  under  the  radical  sign  is 
a  perfect  power  of  the  degree  denoted  by  a  factor  of  the  index. 

Ex.   Reduce  V8  to  its  simplest  form. 

We  have,  -^S=</¥  =  V2  (§  327). 

372.  Case  II.  When  the  expression  under  the  radical  sign  is 
rational  and  integral,  and  has  a  factor  which  is  a  perfect  power 
of  the  same  degree  as  the  surd. 

1.   Reduce  ^54  to  its  simplest  form. 

We  have,  ■^/'El  =  ■\/27^r2  =  ^27  x  V^  (§  165)  =  3-v/2. 

We  can  use  §  165  in  the  above ;  for  we  know  by  §  351  that  it  holds 
when  \/(2,    >/&,  Vc,  •••,  are  surds. 


# 


2.   Reduce  V3  a^b  -  12  d'b'^  + 12  ab^  to  its  simplest  form. 


V3  a%  - 12  a'b^  + 12  ab^  =  V(a'  -  4  a&  +  4  6^)  3  «& 
=  Va2-4a6  +  4  62V3a6  =  (a-2  6)V3a6. 


226       ADVANCED  COURSE  IN  ALGEBRA 

We  then  have  the  following  rule : 

Besolve  the  expression  under  the  radical  sign  into  two  factors, 
the  second  of  ivhich  contains  no  factor  which  is  a  perfect  power  of 
the  same  degree  as  the  surd. 

Extract  the  required  root  of  the  first  factor,  arid  multiply  the 
result  by  the  indicated  root  of  the  second. 

If  the  expression  under  the  radical  sign  has  a  numerical 
factor  which  cannot  be  readily  factored  by  inspection,  it  is 
convenient  to  resolve  it  into  its  prime  factors. 


3.    Keduce  v  1944  to  its  simplest  form. 


-</l944=S/2^  X  3^  =  V2^  x  3«  x  V3^  =  2  x  3  x  V9  =  6V9. 


4.    Eeduce  Vl25  x  147  to  its  simplest  form. 
V125  X  147  =  V5^  X  3  X  7^  =  5  x  7  x  V5^^  =  35Vi5. 

373.  Case  III.  When  the  expression  under  the  radical  sign 
is  a  fraction. 

In  this  case,  we  multiply  both  terms  of  the  fraction  by  such  an 
expression  as  tvill  make  the  denominator  a  perfect  power  of  the 
same  degree  as  the  surd,  and  then  proceed  as  in  §  372. 

/~9~ 
Ex.   Reduce  -^ — ^  to  its  simplest  form. 

Multiplying  both  terms  of  the  fraction  by  2  a,  we  have 

I'V         /9x2a         /"9         ~         I   9  'a^       3      /tt- 

374.  Reduction  of  Surds  of  Different  Degrees  to  Surds  of  the 
Same  Degree. 

Ex.   Reduce  V2,  V 3,  and  V5  to  surds  of  the  same  degree. 

The  lowest  common  multiple  of  2,  3,  and  4  is  12. 

By  §  327,  V2  =  ^2^'  =  ^64. 

^3  =  ^3"^  =  ^8T. 

(         -^5  =  ^^  =  ^125. 


SURDS.    THEORY   OF   EXPONENTS  227 

We  then  have  the  following  rule : 

Make  the  index  of  each  surd  the  L.  C.  M.  of  the  given  indices; 
and  raise  the  expression  under  each  radical  sign  to  a  power  whose 
exponent  is  obtained  by  dividing  this  L.  C.  M.  by  the  given  index. 

The  relative  magnitude  of  surds  may  be  determined  by  reducing  them, 
if  necessary,  to  surds  of  the  same  degree. 

Thus,  since  v^6i  <  '^81  <  ^/UE,  lyollows  that  V2  <  \/3  <  VE. 

ADDITION  AND  SUBTRACTION  OF  SURDS 

375.  To  add  or  subtract  similar  surds  (§  369),  add  or  sub- 
tract their  coefficients,  and  multiply  the  result  by  their  common 
surd  part. 

1.   Required  the  sum  of  V20  and  V45. 
Reducing  each  surd  to  its  simplest  form  (§  372), 
V20  +  V45  =  V4xl  +  V9^o5  =  2  V5  +  3  V5  =  5VB. 


2.   Simplify  ^1  +  ^-^. 

^l+^|-^i-^/^^WF-^/ 


^  x2 
16  ""^ 


MULTIPLICATION  OF  SURDS 
376.   1.   Multiply  V6  by  Vi5. 
By  §  165,  V6  X  Vi5  =  VWxW  =  V90  =  3  VlO. 

2.  Multiply  V2^  by  -^laK 

Reducing  to  surds  of  the  same  degree  (§  374), 

V2^  X  ^/4^  =  ^2^  X  a/4^^  =  -s/2'a^  x  2 V 
=  ^2«a«x2a  =  2a\/2^. 

3.  Multiply  V5by  Vd. 

■V5x-</5  =  </W'x-y/B=V5' 

=  \/5^  (§327)  =^^25, 


228       ADVANCED  COURSE  IN  ALGEBRA 

We  then  have  the  following  rule : 

To  multiply  together  two  or  more  surds,  redtice  them,  if  neces- 
sary, to  surds  of  the  same  degree. 

Multiply  together  the  expressions  under  the  radical  signs,  and 
write  the  result  under  the  common  radical  sign. 

The  result  should  he  reduced  to  its  simplest  form. 


4.   Multiply  3Vl  +  aj-4Va;  by  Vl4-aj  +  2Va;. 


Svl  +  x  — 4Va; 
■yJl-\-x  +  2-yJx 


3(1 +  a5)  +  2Va;  + a;2- 8  05  =  3  -  5  a;  +  2V^  +  ?. 

377.  If  a  surd  is  in  the  form  hVa,  where  a  and  h  are  rational 
expressions,  the  coefficient  may  be  introduced  under  the  radical 
sign  by  raising  it  to  the  ?ith  power,  and  multiplying  the  ex- 
pression under  the  radical  sign  by  the  result. 

Ex.  Introduce  the  coefficient  of  2  a  V3  x^  under  the  radical 
^^^^*     2  aV^^  =  -^8^  ^3^  =  ■V^a^x^x'  =  V^U^. 

378.  A  rational  expression  may  be  expressed  in  the  form  of 
a  surd  of  any  degree  by  raising  it  to  the  power  denoted  by  the 
index,  and  writing  the  result  under  the  corresponding  radical 
sign. 

DIVISION  OF  SURDS 

379.  1.   Divide  a/405  by  -^6. 

By  §  293,  ^405=^405^  ^81  =3^3. 

We  can  use  §  293  in  the  above ;  for  we  know  by  §  351  that  it  holds 
when  Va  and  \/h  are  surds. 

2.   Divide  V^  by  V6. 


SURDS.    THEORY   OF   EXPONENTS  229 

Reducing  to  surds  of  the  same  degree  (§  374), 


We  then  have  the  following  rule : 

To  divide  two  surds,  reduce  them,  if  necessary,  to  surds  of  the 
same  degree. 

Divide  the  expression  under  the  radical  sign  in  the  dividend  by 
the  expression  under  the  radical  sign  in  the  divisor,  and  write  the 
result  under  the  common  radical  sign. 

The  result  should  be  reduced  to  its  simplest  form. 

3.   Divide  VlO  by  V^. 


^40      ■\/40       ^  2^  X  5 
4.  Divide-v/6-2V3by-v/3. 

-v^6  -  2 V3  ^  -y/e      2 A/y  ^  ^/^     o^/g- 
</S  ^3       ^/3 

INVOLUTION  AND  EVOLUTION  OF  SURDS 
380.   1.    Raise  ^12  to  the  third  power. 

{^ny  =  (12^)3  ==  12^  (§  364)  =  12^  =  Vi2  =  2 V3. 

2.  Raise  V2  to  the  fourth  power. 

(^2)* = (2^)^ = 2^ = -s/2' = v'ie. 

Then,  to  raise  a  surd  to  any  power  whose  exponent  is  a 
positive  integer. 

If  possible,  divide  the  index  of  the  surd  by  the  exponent  of  the 
required  power  ;  otherwise,  raise  the  expression  under  the  radical 
sign  to  the  required  power. 

3.  Extract  the  cube  root  of  v^27  x^. 

■V^( V'27^)  =  (a/(3^3)^  =  [(3  x)^-]^  =  (3  x}^  =  ^3^ 


230  ADVANCED  COURSE  IN  ALGEBRA 

4.   Extract  the  fifth  root  of  \/6. 

Then,  to  extract  any  root  of  a  surd, 

If  possible,  extract  tJie  required  root  of  the  expression  under  the 
radical  sign  ;  otherwise,  multiply  the  index  of  the  surd  by  the  index 
of  the  required  root. 

If  the  surd  has  a  coefficient  which  is  not  a  perfect  power  of  the  degree 
denoted  by  the  index  of  the  required  root,  it  should  be  introduced  under 
the  radical  sign  (§  377)  before  applying  the  rule. 

Thus,  v/(4a/2)  =  v/(\/32)  =  V2. 

SPECIAL    METHODS    IN    MULTIPLICATION 

381.  The  rules  of  Chap.  VII  should  be  used  to  find  the  value 
of  any  product  which  comes  under  them. 

1.  Expand  (V6-V3)2. 

By  §  131,  (V6  -  V3)2  =  (V6)2  -  2V6  X  V3  +  (VS)^ 
=  6  -  2  V3^^^  +  3  =  9-  6  V2. 

2.  Expand  (4  +  V^)  (4  -  ^5). 

By  §  132,  (4  +  v'5)(4  -  ■\/5)  =  4^  - (^)2  =  16-^5   (§  380). 

SURD    FACTORS 

382.  The  methods  of  Chap.  VIII  may  be  employed  to 
separate  an  expression  into  surd  factors. 

1.  Factor  2  V^  —  6  a;  by  the  method  of  §  155. 

2V^  -^x  =  2 V^  -  6  (V^)'  =  2V^(1  -  3V«). 

2.  Factor  a  —  6  by  the  method  of  §  171. 

a  -  6  =  ( Va)2  -  (v^)2  =  ( Va  +  V6)  ( Va  -  V6). 
We  may  also  factor  a  -  6  by  the  method  of  §  177  ;  thus, 
a-h  =  {Vly  -  (\/&)8  =  {Va  -  Vb)\_{y/ly  +  V'aVb  +  (v^)^] 
=  ( v^a  -  v^)(\^  +  v^oft  +  v^P). 


SURDS.    THEORY  OF  EXPONENTS  231 

REDUCTION    OF    A    FRACTION    WHOSE    DENOMINATOR    IS 

NOT  A  RATIONAL  EXPRESSION  TO   AN  EQUIVALENT 

FRACTION  HAVING  A  RATIONAL   DENOMINATOR 

383.   Case  I.      JVJieyi  the  denominator  is  a  monomial. 

The  reduction  may  be  effected  by  multiplying  both  terms  of 
the  fraction  by  a  surd  of  the  same  degree  as  the  denominator, 
having  under  its  radical  sign  such  an  expression  as  will  make 
the  denominator  of  the  resulting  fraction  rational. 

Ex.  Reduce  ^  to  an  equivalent  fraction  having  a  rational 
denominator. 

Multiplying  both  terms  by  ^9  a,  we  have 

rj84.  Case  II.  WJieii  the  denominator  is  the  sum  of  a 
rmional  expression  and  a  quadratic  surd,  or  of  two  quadratic 
surds.  _ 

5— V2 

1.  Reduce to  an  equivalent  fraction  having  a  rational 

denominator.  ^H- V2 

Multiplying  both  terms  by  5  —  V2,  we  have 
5_V2^         (5_V2)2 
5  +  V2      (5  +  V2)(5-V2) 

_25j-10V24:2  ,gg.oi    iqnN      27-10V2 

2.  Reduce ^r^ ^~     to  an  equivalent  fraction  having 

2Va-3V^rr6 

a  rational  denominator. 

Multiplying  both  terms  by  2Va  -f  3 Va  —  b, 
3 Va-2 V^^  ^  (3 Va  -  2 V^^^^)  (2 Va  +  3 Vo^) 
2Va-3Va-b     (2 Va  -  3 Vo^^)  (2Va  +  SVcT^) 


^6a-h5VaVa-b-6(a-h)^6b-\-5Va^-ab_ 
4a-9(a-6)  9&-5a 

We  then  have  the  following  rule. 


232      ADVANCED  COURSE  IN  ALGEBRA 

Multiply  both  terms  of  the  fraction  by  the  deyiominator  with  the 
sign  between  its  terms  reversed. 

If  the  denominator  is  the  sum  of  a  rational  expression  and 
two  or  more  quadratic  surds,  or  the  sum  of  three  or  more 
quadratic  surds,  the  fraction  may  be  reduced  to  an  equivalent 
fraction  having  a  rational  denominator  by  repeated  applications 
of  the  above  rule. 

Thus   4-  V3-  V7^  (4-  V3-  V7)(4  +  V3+  V7) 
'  4  +  V3  -  V7      (4  +  V3  -  V7)(4  +  V3  +  VT) 
^  4^-(V3  +  V7/     ,g  ^32x  ^6-2V2T^3-V2l 
(4_^V3)2-(V7>^  12  +  8V3     6  +  4V3' 

Multiplying  both  terms  of  the  latter  by  6  —  4  V3, 

4_V3- V7^(3-V21)(6-4V3) 
4  +  V3-V7  6^-(4V3/ 

^18-6V21-12V3  +  12V7 
-12 

^_9-|-3V21  +  6V3-6V7 
~  6 

The  example  may  also  be  solved  by  multiplying^  both  terms  of  the 
given  fraction  by  4  —  V3  +  V7,  or  by  4  —  V3  -  V7. 

385.  Case  III.  When  the  denominator  is  the  sum  of  a 
rational  expression  and  a  surd  of  the  nth  degree,  or  of  two 
surds  of  the  nth  degree. 

1.   Eeduce  — ^—z  to  an  equivalent  fraction  having  a  rational 
denominator. 
We  have,  (a  +  b)  (a'  -ab-h  b')  =  a'  +  b\ 

Then,  (2  +  ^3)  [2^  -  2  -^3  +  {■^/m  =  2^  +  (^3/. 
Then,  if  we  multiply  both  terms  of  the  fraction  by 
22-2^5+(-\/3)^ 
the  denominator  will  become  irationaL 


SURDS.    THEORY  OF  EXPONENTS  233 

Thus         1    ^^2^-2-^3+_(-C/3)^^4-2-^3  +  a/9 
'  2^V^  2^+  {-y/^f  11 

2.   Reduce  -^ ^  to  an   equivalent   fraction  having  a 

V7  —  V5 
rational  denominator. 

We  have,  (a  -  h)(a^  H-a^ft  +  aW  +  h^)  =  a' -  b\ 

Then,^  if  we  multiply  both  terms  of  the  fraction  by 

i^r  +  {</m<^)  +  (.</^)(</5y  +  {</5y, 

the  denominator  will  become  rational ;  thus, 

-\/343  +  -\/245  4-  a/175  +  A^l^S  . 


2 
The  method  of  Case  III  can  be  applied  to  the  cases  where 
the  denominator  is  of  the  form  V«  +  'Vb,  or  Vot  —  -\/b. 

3.   Reduce  to  an   equivalent   fraction  having  a 

rational  denominator. 

The  lowest  common  multiple  of  the  indices  3  and  2  is  6. 

Now,  (a  -  b)(a'  +  a'b  +  a^b^  +  a'b^-^ab^  +  b')  =a''-  b\ 

Then,  if  we  multiply  both  terms  of  the  fraction  by 
(^'2y  +  {</2y{V5)  +  (■^2)«(  V5)2  +  (a/2)2(V5)3  +  (V2)(^5y 

+  (V5)^ 
the  denominator  will  become  rational. 

Multiplying  both  terms  by  the  above  expression,  we  have 
1         ^2^4  +  2^2V5  4-10  +  5-^4V5  +  25-v/2  +  25V5 
^2-V5  (^2)«-(V5)« 

^  2^44-2^4^^125+10+5v'16a/125+2o-v/2+25V5 
~  4-125 

^     10+2^4+2-^/500  +  5-^2000+25^/24- 25V5 
121 


234  ADVANCED   COURSE  IN   ALGEBRA 

/9  V^ 

4.    Reduce — -   to  an   equivalent   fraction   havinsr  a 

rational  denominator. 

The  lowest  common  multiple  of  the  indices  is  4. 

Now,  (a  +  h)  {p?  -  a'h  +  aW  -  W)  =  a'-  h\ 

Multiplying  both  terms  of  the  fraction  by 

(V2)«  -  ( V2)2(^3)  +  ( V2)(^3)2  -  (^3)^ 
we  have 

V2  -  -^3  ^  ( V2  -  ^3)  (2  V2  -  2  ^3  +  V2  -v/9  -  -</21) 
V2+A/3  (V2)^-(-v/3)^ 

=  4-2V2v'3  +  2^9-V2^27 

-  2  V2  -v/S  +  2  -v^g  -  V2  a/27  +  3 

=  7-4^4^/3  +  4-^9-2-^4-^27 

=  7-4^12  +  4  V3  - 2 -\/l08. 

386.  The  methods  of  §§  383  to  385  are  often  advantageous 
in  finding  the  approximate  value  of  a  fraction  whose  denomi- 
nator is  not  rational. 

Ex.    Find  the  approximate  value  of t=  to  three  places 

of  decimals,  ~~ 

1*  2  +  V2      ^        2-f-V2     24-1.414... 


2-V2      (2-V2)(2+V2)        4-2 


1.707 


387.  In  like  manner,  a  fraction  whose  numerator  is  not 
rational  may  in  certain  cases  be  reduced  to  an  equivalent  frac- 
tion having  a  rational  numerator. 

PROPERTIES   OF  QUADRATIC  SURDS  (§368) 

388.  A  quadratic  surd  cannot  equal  the  sum  of  a  rational 
expression  and  a  quadratic  surd. 

For,  if  possible,  let        Va  =  &  +  Vc, 
where  6  is  a  rational  expression,  and  Va  and  Vc  quadratic  surds. 


SURDS.    THEORY  OF   EXPONENTS  235 

Squaring  both  members,     a  =  6^  +  2  &  Vc  +  c, 
or,  2  6  Vc  z=  a  —  b^  —  c. 

Whence,  V^  =  ^  ~  ^'  ~  ^ 


2b 

That  is,  a  quadratic  surd  equal  to  a  rational  expression. 
But  this  is  impossible ;  whence,  Va  cannot  equal  b  +  Vc. 

389.   7/"  a  +  aA  =  c  +  -Vd,  where  a  and  c  are   rational  ex- 
pressions, and  -y/b  and  -y/d  quadratic  surds,  then 

a  =  c,  and  V&  =  Vd. 

For,  transposing  a,    Vb  =  c  —  a+  Vd. 

Then,  c  —  a  =  0 ;    for,  by   §  388,  a   quadratic   surd   cannot 
equal  a  rational  expression  plus  a  quadratic  surd. 
Therefore,  a  =  c,  and  consequently  V&  =  Vc?. 


390.    If  V  (X  -f  V6  =  V^  +  V^,   w/iere    a,   &,   a;,   and   y   are 
rational  expressions,  and  a  greater  than  y/b,  then 

V  a  —  V^  =  Va7  —  V?/. 
Squaring  both  members  of  the  given  equation, 

a  +  V6  =  X  -{-  2^xy  -\-  y. 
Whence,  by  §  389,        a  =  x-{-y, 
and  -y/b  =  2Vxy. 

Subtracting,         a  —  Vb  =  x  —  2Vxy  +  y. 
Extracting  the  square  root  of  both  members, 


V  a  —  V6  =  V^  —  Vy. 

391.  The  preceding  principles  may  be  used  to  find  the 
square  root  of  certain  expressions  which  are  in  the  form  of 
the  sum  of  a  rational  expression  and  a  quadratic  surd. 

Ex.     Find  the  square  root  of  13  —  VlBO. 


Assume,  Vl3  -  Vl60  =Vx-Vy.  (1 ) 

Then  by  §  390,        V13+VI6O  =  V«  +  V^.  (2) 


236  ADVANCED  COURSE  IN  ALGEBRA 


Multiply  (1)  by  (2),  Vl69  -im  =  x-y. 

Or,  x-y  =  3.  (3) 

Squaring  (1),  13  —  Vl60  =  a;  —  2Vxy  +  ?/• 

Whence,  by  §  389,  x-j-y  =  lS.  (4) 

Adding  (3)  and  (4),  2x  =  16,  or  x  =  S. 

Subtracting  (3). from  (4),        2?/  =  10,  ov  y  =  5. 

Substitute  in  (1),    Vl3 -VIGD  =V8 -V5  =  2  V2-V5. 

392.  Examples  like  that  of  §  391  may  be  solved  by  inspec- 
tion, by  putting  the  given  expression  into  the  form  of  a  tri- 
nomial perfect  square  (§  167),  as  follows : 

Reduce  the  surd  term  so  that  its  coefficient  may  be  2. 

Separate  the  rational  term  into  two  parts  whose  product  shall  he 
the  expression  under  the  radical  sign  of  the  surd  term.  • 

Extract  the  square  root  of  each  part,  and  connect  the  results  by 
the  sign  of  the  surd  term  (§  168). 

1.   Extract  the  square  root  of  8  +  V48. 

We  have,  V48=2Vl2. 

We  then  separate  8  into  two  parts  whose  product  is  12. 
The  parts  are  6  and  2 ;  whence. 


Vs  4- V48  =V6 +  2  Vl2 +  2  =V6+V2. 

2.   Extract  the  square  root  of  22  -  3  V32. 

We  have,         3  V32  =  V9  x  8  x  4  =  2  V72. 

We  then  separate  22  into  two  parts  whose  product  is  72. 
The  parts  are  18  and  4 ;  whence. 


V22  -  3V32  =  V18  -  2V72  -f  4  =  VI8  -  Vi  =  3V2  -  2. 

393.  It  is  sometimes  possible  to  find,  by  the  methods  of 
§§  391  and  392,  the  square  root  of  an  expression  which  is  the 
sum  of  two  quadratic  surds. 

Ex.   Eequired  the  square  root  of  V392  +  V360. 


SURDS.    THEORY    OF   EXPONENTS  237 

V(  V392  +  V360)  =  V[  V2  ( Vi96  +  VlSO)] 


=  V(V2)V14  +  2V45 

=  </2V9  +  2V45  +  5  (§  392) 

=  </2(3  +  V5). 


394.  It  may  be  proved,  as  in  §  390,  that  if  v  a+  V6  =  a;+ V^, 
where  a  and  x  are  rational  expressions,  and  V6  and  Vy  quad- 
ratic surds,  then  3 

V  a  —  Vft  =  a?  —  V^. 

EXERCISE  45 

Reduce  each  of  the  following  to  its  simplest  form : 

1.  \/l21.        3.    y/\mcmK  5.    </m.         7.     7  V112  x^z/^^e. 

2.  v/343.        4.    v^625xiV^.        6.    V^UQ.        8.    v^SO  x  108  x  120. 


9.    V96a'Z)+240«252^150a63.     10.    V(2a:2+a;_i5)(2x2-l9a;+35). 

11.    </!.  12.    -^/^i^.  13.    ^8^^-48x4-72, 

^S  \l6  6c«  ^  3x 

Reduce  to  surds  of  the  same  degree : 

14.    V3,   \/7,  and  Vl5.  15.    ^2^,   v^6"P^,  and  v^iic^ia'. 

16.  Arrange  in  order  of  magnitude  v^S,  \/5,  and  \/il. 

Simplify  the  following : 

17.  V'320  -  \/T35  +  v/625.  18.    J— +a/— +a/-- 

A^18      >'27     ^e 

19.  5^294-9  V150  + 18^/5  _  24  J^. 

\27  \32 

20.  VeSo^P  4-  VTtSo^  +  a62  V63  a'^ft  -  210  a^ft^  +  175  ahK 

21.  (m  +  n)A/?^L±J^  -(m  -  n)4'^^^^=^  -  {mn  -  3  n^)  J— i 

Multiply  the  following : 

22.  VIW-  and  V%h^.  23.    i/M,   -v^,  and  ^. 

'8  '27 

24.    5  Vm  +  %  —  8  y/m  —  n  and  6  y/m  +  n  —  7  Vm  —  «. 
26.   7  V8  +  3  V27  -  2  a/20  and  7  V2  -  3  V3  -  4  V6. 


238  ADVANCED  COURSE   IN   ALGEBRA 

In  the  following,  introduce  the  coefficients  under  the  radical  signs 


6  62 
125  as' 


26.   2v^.  27.  4:x^y^\/3x^y^.  28.  ba^b^-yJ 

29.  «zi^jzz«^zm:.    30. 5'^ii^/23SHiz. 

a  +  X  ^a^  +  2  aic  -  3  a:2  x2  -  2  ^        (x  -  1)2 

Divide  the  following : 
81.    v^lii  by  \/l2.      32.    y/U^  by  v^T^^p.      33.  ^  by  ^^. 

34.    \/2i3  +  \/i8  by  V3.  35.    Vx2  +  2  x  -  3  by  Vx2  -  6  x  +  5. 

36.  \/a2  _  &2  by  \/^2p-qr^p-. 

Simplify  the  following : 

37.  (v/54)3.  39.    (^nx^)5.  41.    v^(V243a^). 

38.  C^^^S"^^)*.  40.    \/(\/l2).  42.    v/(3xv^3^). 
Expand  the  following  by  the  rules  of  Chap.  VIl^: 

43.  (5\/2  +  2V6)2.  47.   \/4  +  2 V3  x  ^^4  -  2 Va. 


44.  (6a/5  +  7V3)(6V5-7V3).    48.  V3  VS  -  2  V?  x  V3  V5  +  2  ^7. 

45.  (3\/x  +  ?/-4\/x-2/)2.  49.   ( Va  +  Vft  +  Vc)  (  Va  +  Vd  -  Vc) . 

46.  (3V5-2>/lO)3.  50.  (v^  +  v^9)(  v^  -  v^). 

51.  (2V3  +  5\/2-V5)(2\/3-5\/2+\/5). 
52.  (\/lO-4V5  +  5V2)2.  53.   (2V2  +  V6 -VS)^. 

Factor  the  following : 

54.  V2a  +  \/3a.       55.  x-\/x-20.       56.  ac  ■{■  ay/d  -  cVb -Vbd. 

57.  Factor  v^  -  v^i  by  the  rule  of  §  171. 

58.  Factor  a  —  6  by  the  method  of  §  178,  taking  Va  —y/b  for  the  first 
factor. 

Reduce  each  of  the  following  to  an  equivalent  fraction  having  a 
rational  denominator  : 


65. 


59.  __^«!_.  62.  V^^Ty^-y/^^^^^ 

v^27  ab^c^  '  Vx2  +  y^  +  Vx'^  -  ?/2  ""'    </a  +  y/b 

60     Vx  +  Vj  63. ^ _       •  66.    ^. 

Vx-Vy  VVll +  3 -VVll-3  m-y/n 

gj     3V5-V3  g4  V6  +  V3-3V2  g^           1 


4V6+5V3  V6-V3  +  3\/2  v^  +  v^ 


SURDS.     THEORY   OF   EXPONENTS  239 


1  o„      1  70  n-vs 


3V7- 

5V3 

78. 

V343  +  Vl68. 

79. 

V1058 

-  V896. 

S.    (x  +  V^)V  S3.    (Vf-lV^y.      B4.    (1|  +  ^,. 


3  +  V2  ^a-v^6  ^4  +  V3        • 

Eind  the  approximate  value  of  each  of  the  following  to  five  places  of 
decimals  : 

71.         5  72.    ^-^5  „o    4V7  +  7V3 

'    3-V3  '    V6+V5 

Extract  the  square  root  of  each  of  the  following 

74.  24  +  2\/l4d.  76.    38-5V52. 

75.  87-V2240.  77.   61 +28  VS. 
80.   2x-Sy-2Vx^-3xy.  81.  4  a  +  2  +  2V3  a^  +  8  a  -  3. 

Expand  by  the  Binomial  Theorem  : 

Find  the 
85.   5th  term  of  (2x-\-  SVyy^  86.    9th  term  of  (a/-  -  -^-V^- 

SOLUTION  OF  EQUATIONS   INVOLVING  THE  UNKNOWN 
NUMBERS  UNDER  RADICAL  SIGNS 

395.  To  solve  an  equation  involving  the  unknown  numbers 
under  radical  signs,  we  transpose  the  terms  so  that  a  surd  term 
may  stand  alone  in  one  member,  and  then  raise  both  members 
to  a  power  of  the  same  degree  as  the  surd. 

If  surd  terms  still  remain,  we  repeat  the  operation. 

396.  We  will  now  prove  that 

,^  If  both  memlmrs  of  an  equation  be  raised  to  the  same  positive 
integral  /^OT^Hle  resulting  equation  ivill  have  all  the  solutions 
of  the  given  equation,  and,  in  general,  additional  ones. 

Consider  the  equation       A=  B.  (1) 

Raising  both  members  to  the  nth  po^er,  n  being  a  positive 
integer,  we  have     ^„  ^  ^^^  „,  ^„  _  B«  =  0.  (2) 

Factoring  the  first  member  (§  178), 

(A  -  B)  {A--'  +  A^-'B  +  •  •  •  4-  ^-')  =  0.  (3) 


240       ADVANCED  COURSE  IX  ALGEBRA 

By  §  182,  (3)  is  equivalent  .to  the  equations 

I  A-B  =  0,  OT  A  =  B,  and 

I  J.«-i  4-  A^-'B  +  •••  +  B'^-^  =  0. 

Thus,  equation  (2)  has  not  only  the  solutions  of  (1),  but  also 
the  solutions  of 

Jn-l  _|_  Jn-2^  +  ...  +  J5"-l  =  0, 

which,  in  general,  do  not  satisfy  (1). 
Take,  for  example,  the  equation 

x  =  3.  (1) 

Squaring  both  members,  we  have 

5c2  =  9^  or  a;2  -  9  =  0.  (2) 

Factoring  the  first  member,  and  placing  the  factors  separately  equal  to 
0(§  182),  we  have  ^  +  3  =  0,  or  x  =-3; 

and  ic  —  3  =  0,  or  a:  =  3. 

Thus,  equation  (2)  has  the  root  3,  and,  in  addition,  the  root  —  3. 

397.  It  follows  from  §  396  that  all  solutions  obtained  by 
raising  both  members  of  an  equation  to  any  positive  integral 
power  should  be  verified;  only  such  as  satisfy  the  given  equation " 
should  be  retained. 

In  verifying  solutions  of  equations  involving  the  unknown 
numbers  under  radical  signs,  it  should  be  carefully  borne 
in  mind  that  only  ^principal  values  of  the  roots  are  ■  considered 
(§  162). 

398.  Examples. 


1.    Solve  the  equation  V ar^  —  5  —  a;  =  —  1. 
Transposing  —  x,  Va^  —  5  =  x  —  l. 

Squaring  both  members,         x^  —  5  =  x^  —  2  x  -\- 1.    ■ 
Transposing,  ^     2  x=:6;  whence,  x  =  3. 

Putting  aj  =  3,  the  given  first  member  becomes 

V9^^-3  =  2-3  =  -l. 
"Thus,  the  solution  ic  =  3  satisfies  the  given  equation. 


SURDS.    THEORY  OP  EXPONENTS  241 

2.    Solve  the  equation  V2  x  —  1  -f-V2a;  +  6  =  7. 


Transposing  ■V2x  —  1,  ^2x-i-6  =  7  —  V2  a;  —  1. 


Squaring,  2  ic  +  6  =  49  -  14  V2  a;  -  1  +  2  a;  -  1. 

Transposing,      14  V2  a;  —  1  =  42,  or  V2x  —  1  =  3. 

Squaring,  2  a;  —  1  =  9 ;  whence,  x  =  5. 

Putting  a;  =  5,  the  given  first  member  becomes  V9  +  Vl6  =  3  +  4  =  7. 
Thus,  the  solution  a;  =  5  is  correct. 

_  o 

3.   Solve  the  equation         Va;  —  6  —  Va; 


Va;-6 


Clearing  of  fractions,  x  —  6  —  Vx'^  —  6  a;  =  3. 


Transposing,  —  Vx^  —  ()x=9  —  x. 

Squaring,  a^  —  6  a;  =  81  —  18  a;  +  a;l 

Then,  12  a?  =  81 ;  whence,  x  =  ~  =  —. 

12      4 

27 
Putting  X  =  — ,  the  given  first  member  becomes 

4 


V!-A/f=|^-f^s=-^- 


2 

■  q  q 

The  second  member  becomes  -^—  = =  2  VS. 


\4      2 


fi      2 

27 
Thus,  the  solution  a:  =  —  does  not  satisfy  the  given  equation  ;  in  this 

case  there  is  no  solution. 

4.    Solve  the  equation  V2  —  3  a;  +  Vl  +  4a;  =  VS  +  x. 
Squaring  both  members, 

2  -  3  a;  +  2  V2  -  3  a;Vl  +  4  a;  +  l  +  4a;  =  3  +  a;. 


Whence,  2V2-3a;Vl +  4a;  =  0; 


or,  V2-3a;Vl  +  4a;  =  0. 

Squaring,  (2  -  3  a?)  (1  +  4  a;)  =  0. 

Solving  as  in  §  182,       -  '  2-3  a;- 0,  or  a;  =  ?;' 

o 

and^  1  +  4  a;  =  0,  or  a;  =  —  i  • 

Both  values  satisfy  the  given  equation. 


5 


242  ADVANCED   COURSE  IN   ALGEBRA 

EXERCISE  46 

Solve  the  following : 

1.  ^8 :«3_36 0:^  +  3  =  2:..  3.     ^^^+1+^^^! 

V3  X  +  1  -  V3  X     4 

2.  V5a:-19-\/5a:+14=-3.       4.    Vo^ -V6(C- 11  = ^ 

V6  ic  - 11 

5.  Vax  +  &c  +  Vax  —  hc  =  V4  ax 

6.  \/a;-2a-Vx=         ^     -• 

Vx  —  2  a 

8.  Vx2  -  5  X  -  2  +  Vx^  +  3a;  +  6  =  4. 

9.  \/4x+  1  -  Vx-8  =  \/9 X  -  83. 

10.  V2  X  -  5  a  +  V3  X  +  4  6  =  V5  x  -  5  a  +  4  &.     . 

11.  V(x  +  a) (x  +  6)  +  V(x  -  a)(x-&)  =  V2x2  +  2a6. 

12.  V2x+  5  +  V3x-2  =  V(5  x  +  3  +  V24x2,+  15). 
.j3  3\/2x-  1  +  4  ^  V2  X  -  1  +  6  ^ 

6\/2x-  1-1      2\/2x-  1  -5 

14.  Vx2  +  10  X  -  24  +  Vx2  +  7x  +  12  =  \/4  x^  +  17  x  +  4. 

15.  \/2  X  +  1  +  \/3  X  +  2  =  Vx+  2  +  V4x+1. 

IRRATIONAL  NUMBERS 
399.   Consider  the  series 

and  Oj,  a2j  •••,  a^,  ••• ;  (2) 

in  which  the  terms  of  (1)  continually  decrease,  and  the  terms 
of  (2)  continually  increase;  and  let  a\  —  a^  approach  the  limit 
0,  when  r  is  indefinitely  increased. 

Then,  any  expression  which  is  not  a  rational  number,  and 
which  is  greater  than  the  terms  of  (1),  and  less  than  the  terms 
of  (2),  is  called  an  Irrational  Number. 

The  common  limit  of  the  rth  terms  of  (1)  and  (2),  when  r  is 
indefinitely  increased,  is  considered  the  value  of  the  above 
irrational  number. 

A  surd  is  one  form  of  irrational  number. 

Rational  and  irrational  numbers  are  called  Real  Numbers. 


SURDS.    THEORY  OF   EXPONENTS  243 

400.  Consider,  for  example,  the  expression  of^\ 
a  ^  lies  between  the  two  series  (compare  §  343), 

a\  a»  a}-^,  ■■.,  (1) 

and  a},  a}*,  a'",  •••.  (2) 

n 

If  we  represent  the  rth  term  of  (2)  by  aio*""^,  the  rth  term 
of  (1)  will  be  a?^\ 

Then,  the  difference  between  the  rth  terms  will  be 

n+l  w  n  1 

dic-i  _  aio'-i  =  ai(F-i  (ai(F-i  _  1).  (3) 

n 

Now,  aio'-i  is  always  less  than  al 

1 

And,  by  §  343,  aio^^  approaches  the  limit  <  or  1  (§  359), 
when  r  is  indefinitely  increased. 

Therefore,  the  expression  (3)  approaches  the  limit  0  when  r 
is  indefinitely  increased. 

Hence,  a^^  is  the  limit  of  the  ^-th  term  of  either  (1)  or  (2), 
when  r  is  indefinitely  increased. 

A  meaning  similar  to  the  above  will  be  attached  to  any  form 
of  irrational  exponent. 

401.  The  definitions  of  Addition  and  Multiplication,  given 
in  §  348,  hold  when  any  or  all  of  the  numbers  involved  are 
irrational; 

402.  It  may  be  shown,  as  in  §  350,  that  the  fundamental 
laws  of  §§  12  and  14  hold  when  any  of  the  letters  involved 
represent  irrational  numbers. 

Then,  every  statement  or  rule,  in  the  remaining  portions  of 
Chap.  I,  or  in  Chaps.  II  to  XVI,  inclusive,  in  regard  to 
expressions  where  any  letter  involved  represents  any  rational 
number,  holds  also  when  the  letter  represents  any  real  number. 

Also,  the  theorems  of  §§  165  and  293  may  be  jjroved  to  hold 
when  any  or  all  of  the  letters  a,  6,  c,  etc.,  represent  irrational 
numbers  which  are  positive  if  n  is  even ;  and  the  theorem  of 
§  327  may  be  proved  to  hold  when  a  is  any  irrational  number 
whose  mth  power  is  positive  if  n  is  even. 


244  ADVANCED   COURSE  IN  ALGEBRA 

403.  We  will  now  prove  that  equation  (1),  §  356,  holds  when 
m  is  a  positive  rational  number,  and  n  a  surd  of  the  form  Vb, 
where  p  and  h  have  the  same  meanings  as  in  §  348. 

By  §  399,  a*"  x  a^*  is  the  limit,  when  r  is  indefinitely  in- 
creased, of  cC^  X  a*'-,  where  h^  has  the  same  meaning  as  in  §  348. 
p  - 
Also,  a'"+^*   is  the   limit   of   a'^^^  when   r  is   indefinitely 

increased. 

But  since  m  and  h^  are  rational, 

a"*  X  aJ'rz^a'^+^r  (§§  85  or  362). 

Then,  oT  X  a**-  and  a'"+*'-  are  functions  of  r  which  are  equal 
for  every  rational  value  of  r;  and,  by  §  252,  their  limits  when 
r  is  indefinitely  increased  are  equal. 

Hence,  a"*  x  a^^=  a'^+^^ 

In  like  manner,  we  may  prove 

a'"  X  a'*  =  a^^+^j 

in  every  case,  not  previously  considered,  where  a,  m,  and  n  are 
any  real  numbers,  provided  a™  and  a"  are  real  numbers. 

404.  It  may  be  proved,  as  in  §  403,  that 


=  a" 


a"" 
and  (a&c...)'^  =  a"&V..., 

in  every  case,  not  previously  considered,  where  a,  6,  c,  •••,  m, 
and  w  represent  any  real  numbers,  provided  a*",  a"*,  a*"",  6", 
c%  •••,  are  real  numbers. 

405.  It  follows  from  §§  402  to  404  that  every  result  in 
§§  367  to  398  inclusive  holds  when  any  letter  involved,  except 
when  the  index  of  a  root,  represents  any  irrational  number, 
such  that  every  expression  of  the  form  ^a  is  an  irrational 
number. 


SURDS.    THEORY  OF  EXPONENTS  245 

GRAPHICAL  REPRESENTATION  OF  IRRATIONAL  NUMBERS 

406.  It  was  shown,  in  §  343,  that  V2  was  intermediate  in 
value  between  the  series 

2,  1.5,  1.42,  ...,  and  1,  1.4,  1.41,  ...; 

and  that  V2  is  the  limit  of  the  Hh  term   of  either   series 
when  r  is  indefinitely  increased. 

■" 1 i \ — i — \ — 1 — \ '1 

0  A  Ai     Az     P     -Bz     -^i  -^ 

Let  A,  ^1,  Ao,  '",  A^,  •••  be  the  points  in  the  scale  of  §  57, 
corresponding  to  the  numbers  1,  1.4,  1.41,  •••;  and  B,  B^, 
B2,  •••,  B^,  '•'  th€  points  corresponding  to  the  numbers  2,  1.5, 
1.42,  .... 

The  distance  between  A^  and  B^  approaches  the  limit  0, 
when  r  is  indefinitely  increased ;  that  is,  OA^  and  OBr  approach 
the  same  limit. 

If  OP  is  this  limit,  OP  represents  V2. 

In  like  manner,  a  point  exists  whose  distance  from  0  repre- 
sents any  irrational  number  whatever. 


246  ADVANCED  COURSE  IN   ALGEBRA 

XVIII.     PURE    IMAGINARY    AND    COMPLEX 
NUMBERS 

It  will  be  understood,  in  §§  407  to  414,  inclusive,  that  every  letter 
represents  a  positive  real  number  (§ 


PURE  IMAGINARY  NUMBERS 

407.  We  define  S/—  a,  where  a  is  any  positive  real  number, 
and  n  an  even  positive  integer,  as  an  expression  whose  nth 
power  equals  —  a. 

That  is,  (^■^Z~ay  =  -a. 

The  symbol  -yZ—ais  called  a  Pure  Imaginary  Number. 

It  is,  of  course,  impossible  to  find  any  real  number  whose 
nth.  power  equals  —  a ;  but  there  are  many  advantages  in  in- 
cluding in  the  number-system  of  Algebra  the  result  of  any  finite 
number  of  the  operations  addition,  subtraction,  multiplication, 
division,  involution,  and  evolution,  with  rational  numbers. 

The  pure  imaginary  number  V—  1  is  called  the  imaginary 
unit;  it  is  usually  represented  by  the  letter  i. 

OPERATIONS  WITH  PURE   IMAGINARY  NUMBERS 

408.  In  deriving  the  rules  for  operations  with  pure  imaginary 
numbers,  we  shall  follow  the  method  employed  in  Chap.  II ; 
that  is,  we  shall  assume  that  the  fundamental  laws  of  §§  12  and 
14  hold  for  such  numbers,  and  find  what  meaning  must,  in  con- 
sequence, be  attached  to  the  operations.     (Compare  §  50.) 

It  follows  from  this  that  every  statement  or  rule  in  the 
remaining  part  of  Chap.  I,  or  in  Chaps.  II  to  XVI,  inclusive, 
in  regard  to  expressions  where  any  letter  involved  represents 
any  rational  number,  holds  equally  where  the  letter  represents 
a  pure  imaginary  number.     (Compare  §  351.) 

409.  To  prove  that,  if  a  is  any  positive  real  number, 

V—  a  =  Va  V—  1. 
By  §407,  (^-—ay  =  -a.  (1) 


PURE   IMAGINARY  AND  COMPLi:X  NUMBERS     247 


And  since  the  result  of  §  129  holds  when  any  of  the  letters 
a,b,c,  •••  represents  a  pure  imaginary  number  (§  408), 

=  ax(-l)  =  -a.  (2) 

V— a  =  Va  V—  1. 


From  (1)  and  (2), 
Then  by  §  163, 

410.   By  §409, 


V327  =  V27V'^ 

=  3V3V^1  (§372) 

=  3V^^  (§409). 

It  is  evident  from  this  that  the  methods  of  §§  372,  373,  and 
377  hold  for  pure  imaginary  numbers. 

411.  Powers  of  V—  1. 

By  §407,  (V=T)2  =  -1. 
Then, 

(V3i)3  =  (^^ly  X  V^=i:  =  (- 1)  X  V"^  =  -  V^=l; 

(V^=i:)*=  (V^^^y  X  (V^2  =  (-1)  X  (- 1)  =  1 ;  etc. 
In  general,  if  n  is  any  -positive  integer, 

( V3i)4n  _  |-(  viri;)4]n  =  1-  _  1 . 

( V^^)'''+^  =  ( V^)^*^  X  V^I = V^=i: ; 

( V^^)*"+2  =  (V^l)^'^  X  (V^^y  =  (■y/'^y  =  - 1 ; 

(V^=3)'"+« = ( v^=^)^"  X  (V^^y = (V^f = -  v^=o[.    . 

412.  Addition  and  Subtraction. 

Two  pure  imaginary  numbers  may  be  added  or  subtracted 
by  the  method  of  §  375. 


Ex^  Add  V -  4  and  V-M. 


*By  §  410, 


4  +  V-Tse  =  2  V^n  +  6  V- 1 

=  (2  +  6)  V^n:  (§  14,  III)  =  8  V^T. 


248 


ADVANCED  COURSE   IN  ALGEBRA 


413.  Multiplication. 

The  product  of  two  or  more  pure  imaginary  numbers  may 
be  obtained  by  aid  of  the  principles  of  §§  409  and  411. 

1.  Multiply  V3r2  by  V^^. 

By  §409,  V^^xV^^  =  V2V^^xV3V?nri 

=  V2  V3  ( V"^)^  %  §  14, 1  and  II, 
=  V6  X  (-1)  (§  411)  =- V6. 

2.  Multiply  together  V^^,  V-16,  and  V-25. 
V^^  X  V^=l6  X  V^^^  =  3  V^^  X  4  V^  X  5  V^^. 

-  =  60  (V^=l)^  =  60  (-  V^I)  (§  411)  =  -  60 V^:i. 

3.  Multiply  2 V^^  +  V^=^  by  V^^  -  SV~^. 

Since  all  the  rules  of  Chap.  IV  hold  for  pure  imaginary 
numbers,  we  can  multiply  as  in  §  88. 

2(-2)  +    V2V5(V^^ 

^     -6V2V5(V:=1)2_3(_5) 

- 4  - 5VlO(- 1)  + 15  =  11  +  5VlO. 

414.  Division. 

1.  Divide  V^^^IO  by  V^^. 

By  §409,   V^^V40V^^V40^^3^2V2. 
V-5       V5V-1       V5 

Since  the  rule  of  §  96  holds  for  pure  imaginary  numbers, 
V— 1  is  cancelled  in  the  same  manner  as  a  real  factor. 

2.  Divide  Vl5  by  V^=^. 
By  §  411, 

-:^=-V^^^^)^=-V5(V:=T)  =  -Vi:5. 

V=3  V3V^^ 


PURE   IMAGINARY  AND  COMPLEX  NUMBERS     249 

0~.fo  Q     /       T) 

3.   Keduce to  an  equivalent  fraction  having 

2V3+3V-2 
a  real  denominator. 

This  may  be  effected  by  multiplying  both  terms  of  the 
fraction  by  the  denominator  with  the  sign  between  its  terms 
changed. 

Multiplying  both  terms  by  2  V3  —  3  V^^, 

•2V3-3V^:2  ^    (2V3-3y^=:^y     .^  ^^^. 
2  V3  +  3  V^^      (2  V3)2  -  (3  V^2 

^12-12V3V2V^^  +  9(-2)  ,.  .o-,x 
12_9(-2)  ^^        ^ 

^_6-12V6V^:ri^     l4,2V3r6 
30  5 

EXERCISE  47 

In  the  following  examples,  every  letter  occurring  under  a  radical  sign 
is  supposed  to  represent  a  positive  real  number. 

1.  What  is  the  value  of  (^y/^^y^  ?    of  (a/^)3i  ?    of  (^V^y^  ? 
Simplify  the  following : 

2.  (V^^)6.  3.    (\/^5)7.  4.    5V^  +  2V^^. 


5. 

7V" 

-80-^ 

\V^ 

-125 

-V' 

-245 

%. 

6. 

V- 

-H- 

27 
16' 

-V- 

49 
"12" 

-i- 

16 
27' 

• 

Multiply  the  following : 

7. 

v^ 

-  5  and 

V^ 

45. 

9. 

v^ 

-  6  and  y/h  — 

a. 

8. 

vC 

ry,  v^ 

"14, 

and  V^ 

21. 

10. 

V- 

6,   V-8,  and 

VlO. 

11. 

v^ 

-18,    V- 

-27 

,  V- 

-32, 

and 

v~ 

48. 

12.  8  +  5  V^^  and  7  -  6\/^^. 

13.  6V^^  -  5a/^=^  and  4^/^  +  SV^^. 
Divide  the  following : 

14.  \/^^n08  by  >A^.     16.   \/l20  by  V^.  18.   Vsi  by  V^^, 

15.  \/^41  by  V^,      17.    V^:^^224  by  V^.      19.    V^^^lGO  by  VS. 


250       ADVANCED  COURSE  IN  ALGEBRA 

Expand  the  following  by  the  rules  of  Chap.  VII : 

20.  (4  V-~3  +  5a/^^)2.     22.   (3V^=T  +  2V^^)(3\/^^  -  2V^^). 

21.  (v^ir6-\/^^)3.         23.   (3\/^^  +  V^^-2\/^^)2. 

24.    (V3^-v/^r5  +  V'^^)(v'^+V^^-V"^). 

Reduce  each  of  the  following  to  an  equivalent  fraction  having  a  real 
denominator : 

25.    ^-^^.         26.    2x/^  +  7x/i:3.         ^^^  1 


34-V-7  4\/-5-3>/^^  2  +  3\/-2-\/-6 

Expand  the  following  by  the  Binomial  Theorem  : 
28.    (V^^  +  SV'^2y.  29.    iVa-y^^y, 

Simplify  the  following  : 

o^              3            ,            6                   oi     2\/5  +  3>Ar6      2\/5-3\/^ 
au.    — — —  -| •        oi.   — — —^^ • 

5_4V-5      7  +  2V^  2V5-3a/^     2V5  +  3V-6 

COMPLEX  NUMBERS 

It  will  be  understood,  throughout  the  remainder  of  the  present  chapter, 
that  every  letter  represents  a  real  number. 

415.  The  expression  a  -i-bi  (§  407),  where  a  and  b  are  any- 
real  numbers  whatever,  is  called  a  Complex  Number. 

In  operations  with  complex  numbers,  we  shall  assume  the 
laws  of  §§  12  and  14  to  hold. 

It  follows  from  this  that  the  statement  in  the  last  paragraph 
of  §  408  is  equally  true  of  complex  numbers. 

416.  Addition,  Subtraction,  Multiplication,  and  Division  of 
Complex  Numbers. 

1.  Add  a-{-  bi  and  c  +  di. 

Since  the  laws  of  §  12  hold  for  complex  numbers, 

(a  +  bi)  +  (c  +  cZz)  =  a  +  c  4-  bi  +  di 

=  a  +  c  +  (b  +  d)i,  by  §  12,  II,  and  §  14,  III. 

2.  Subtract  c  +  di  from  a  +  bi. 

(a  4-  bi)  -  (c  +  di)  =  a -^bi  -  c  -  di  (^  81) 

=  a  -  c  +  (6  -  d)i,  by  §  12,  and  §  24,  (7). 


PURE   IMAGINARY  AND  COMPLEX  NUMBERS     251 

3.  Multiply  a-\-bihj  c  +  di. 

By  §  88,  (a  +  bi){c  +  di)  =  ac-{-  adi  +  bci  +  Mi^ 

—  ac  —  bd-{-  {ad  -f-  bG)i,  by  §  411. 

4.  Express  the  quotient  of  a  +  bi  by  c  +  di  as  a  complex 
number. 

Multiplying  both  numerator  and  denominator  by  c  —  di,  we 

have 

a  +  ^<^'  _  (g  +  &t)(c  —  di)  _  ac  —  adi  +  bci  —  6d^^ 

c  +  di     (c  +  di)(c  —  di)  &  —  dH^ 

_ac-\-bd-{-(bc—ad)i_ac-{-bdbG  —  ad. 
TTd'  ~  c'-\-d'       c^-\-d^^' 

417.  It  follows  from  §  416  that  the  result  of  any  finite  num- 
ber of  additions,  subtractions,  multiplications,  and  divisions, 
performed  upon  complex  numbers,  is  a  complex  number. 

418.  By  the  definition  of  0, 

1X0  =  i{a  -  rt)  =  ia  -  m(§  24,  (7))  =  0. 

It  follows  from  this  that  the  complex  number  a  -|-  bi  becomes 
a  real  number  when  6  is  0. 

It  also  becomes  a  pure  imaginary  number  when  a  is  0. 
Hence,  a  +  6i  cannot  equal  0  unless  both 

a  =  0,  and  6  =  0. 

419.  Ifa-^bi  =  c-\-  di,  where  a,  b,  c,  and  d  are  real  numbers, 
then  a  =  c,  and  b==d. 

Transposing  the  terms  of  the  given  equation,  we  have 

a-~c-\-{b  —  d)i  =  0. 
Then,  by  §418,    a- c  =  0,  and  6-d  =  0. 
Whence,  a  =  c,  and  b  =  d. 

420.  Square  Root  of  a  Complex  Number. 


We  will  now  prove  that  Va  -h  bi,  where  a  and  b  are  re^,!, 
can  be  expressed  in  the  form  -y/x  +  i  V^,  where  Va;  and  Vy 
are  real. 


252  ADVAJ^CED   COURSE  IN   ALGEBRA 

Squaring  both,  members  of  the  equation 

Va  +  M  =  Va?  +  i  V^,  (1) 

we  have  a-^bi  =  x-\-2i  -Vxy  —  y. 

Then  by  §  419,  a^x-y,  (2) 

and  6i  =  2  1  w'xy. 

Subtracting,  a  —  hi  =  x  —  2i  -Vxy  —  y. 

Extracting  the  square  roots  of  both  members, 

Va  —  hi  —  -Vx  —  i  -Vy.  (3) 

Multiply  (1)  by  (3),  -V^Tb'  =  x  +  y.  ■     (4) 

Add  (2)  and  (4),  y/W+h^  +  a^2x,ovx  =  ^^^^^'  +  ^. 


Subtract  (2)  from  (4),  Va^  +  6^  —  a  =  2  ?/,  or  ?/  = 


Va'  +  &'  -  a 


It  is  evident  from  this  that  ■\/x  and  V^/  are  real. 

421.  It  follows  from  (1)  and  (3),  §  420,  that 

If  Va  H-  hi  =  Va;  +  i  V?/,  w/iere  a,  6,  a;,  and  y,  cure  real  num- 
hers,  then  Va  —  hi  =  Va;  —  z  V?/- 

422.  The   preceding   principles   may  be  used  to  find  the 
square  root  of  a  complex  number. 

1.    Find  the  square  root  of  7  —  6  V— 2. 


Assume,  V7- 6V-2  =V^- V^V^^.  (1) 

Then  by  §  421,     Vt+W^  =  V^^  +  V^ V^^.  (2) 

Multiplying  (1)  by  (2),  we  have 


V49-36(-2)  =  a;  +  2/, 
or,  x-\-y  =  ll.  (3) 

Squaring  (1),  7  —  6 V^^  =  x  —  2 V^V—  1  —  2/. 

Whence  by  §  419,  x  —  y  =  7.  (4) 

Add  (3)  and  (4),  2  a;  =  18,  or  a;  =  9. 

Subtract  (4)  from  (3),  2  3/  =  4,  or  2/  =  2. 


PURE  IMAGINARY  AND   COMPLEX  NUMBERS     253 


Substitute  in  (1),  V7  -  6-V-2  =  V9  -  V2 V- 1=3-  V^. 

The  example  may  also  be  solved  by  the  method  of  §  392. 

We  have  6  V^  =  2V9  x(-2). 

We  then  separate  7  into  two  parts  whose  product  is  9  x  (—  2). 

The  parts  are  9  and  -2;  then,  V7-6V^  =  V9-\/^=3- \^^. 
We  may  also  find  by  the  above  methods  the  square  root  of  an 
expression  of  the  form  a  +  V6,  or  a  —  V6,  when  a  is  negative. 

2.   Find  the  square  root  of  —  35  + 12  V6. 
We  have  V_  35  + 12 V6  =  V^l ^35 -  2 V216. 
Separating  35  into  two  parts  whose  product  is  216, 
V_35  +  12V6  =  V^^ V(27  -  2 V27^  +  8) 

=  v^ri(  V27  -  V8) = 3  V^=^-  2  V"=^. 

EXERCISE  48 

Extract  the  square  roots  of  the  following : 


1.   13  +  V^^l92.  3.    -  38-  8 v^-^30.  5.    -  52  -  2 V640. 


2.    18  -  5V'^^28.  4.    -  26  +  V480.  6.   2 V-  16. 

423.  Putting  a  =  0  and  6  =  1,  in  §  420,  we  have  x  =  y  =  - 
Substituting  these  values  in  (1),  we  have 

424.  Cube  Root  of  a  Complex  Number. 


We  will  now  prove  that  if  Va  -\-  bi  =  c-{-  di,  where  a,  &,  c, 
and  d  are  real,  then  -Va  —  bi  =  c  —  di. 
Cubing  both  members  of  the  equation 

Va  -\-bi  =  c-\-  di, 
we  have,  by  §  411,  a  +  bi  =  c^ -\- 3  c'di  -  S  cd^  -  dH. 

Then  by  §  419,  a  =  c^-S  cd^ 

and  bi  =  3  cHi  —  dH. 

Subtracting,        a—bi  =  (?  —  3  &d%  —  3  cd^  +  d?L 


254  ADVANCED   COURSE   IN  ALGEBRA 

Extracting  the  cube  root  of  both  members, 


V  a  —  bi  =  c  —  di. 

425.  The  complex  numbers  a  +  hi  and  a  — hi  are  called 
Conjugate. 

We  have    {a  +  hi)  x  (a  -  hi)  =  o?-  hH\  =  a^  +  h\ 
Also,  (a  +  hi)  +  (a  —  hi)  =  2  a. 

Hence,  the  sum  and  product  of  two  conjugate  complex  numbers 
are  real. 

It  will  be  shown  in  the  Appendix  (§  804),  that  any  even  root  of  a 
negative  number,  or  any  root  of  a  pure  imaginary  or  complex  number, 
can  be  expressed  as  a  complex  number. 

GRAPHICAL  REPRESENTATION  OF  PURE  IMAGINARY  AND 
COMPLEX  NUMBERS 

426.  Let  XX'  be  a  fixed  straight  line,  and  0  a  fixed  point 
in  that  line. 

It  was  shown  in  §§57  and  406  /^"^ 

that  any  positive  real  number,  +  a,     X    ^/   -^    o  -ta    A     ^ 
could   be   represented   by   the  dis- 
tance from  0  to  ^,  a  units  to  the  right  of  0  in  OX ;  and  any 
negative   real   number,   —  a,  by  the   distance   from    0  to  A', 
a  units  to  the  left  of  0  in  OX'. 

427.  Since  —  a  is  the  same  as  (-f  a)  x  (—  1),  it  follows  from 
§  426  that  the  product  of  -f  a  by  —  1  is  represented  by  turning 
the  line  OA  which  represents  the  number  +  a,  through  two 
right  angles,  in  a  direction  opposite  to  the  motion  of  the  hands 
of  a  clock. 

We  may  then  regard  —  1,  in  the  product  of  any  real  number 
by  —  1,  as  an  operator  which  turns  the  line  which  represents 
'\Mq  first  factor  through  two  right  angles,  in  a  direction  opposite 
to  the  motion  of  the  hands  of  a  clock. 

428.  Graphical  Representation  of  the  Imaginary  Unit. 

By  the  definition  of  §  407,  —  1  =  ^  x  i. 


PURE   IMAGINARY  AND   COMPLEX   NUMBERS     255 


Y 

■B 

C- 

+ai 

+i 

^    . 

Y 

-i 

0-^cb   A 

C] 

-ai 

■b' 
y' 

being 


Then,  since  one  multiplication  by  i, 
followed  by  another  multiplication  by  i, 
turns  the  line  which  represents  the  first 
factor  through  two  right  angles,  in  a  direc- 
tion opposite  to  the  hands  of  a  clock,  we 
may  regard  multiplication  by  i  as  turning 
the  line  through  one  right  angle,  in  the 
same  direction. 

Thus,  let  XX'  and  YY'  be  fixed  straight 
lines,  intersecting  at  right  angles  at  0, 
arranged  as  in  the  figure  of  §  270. 

Then,  if  +a  be  represented  by  the  line  OA,  where  ^  is  a 
units  to  the  right  of  0  in  OX,  -f  ai  may  be  represented  by 
OjB,  and  —  ai  by  OB'^  where  ^  is  ci  units  above,  and  B'  a 
units  below,  0,  in  YY\ 

Also,  4-  i  may  be  represented  by  OC,  and  —  i  by  OC,  where 
C  is  one  unit  above,  and  O  one  unit  below,  0,  in  YY\ 

It  will  be  understood  throughout  the  remainder  of  the  chapter  that, 
in  any  figure  where  the  lines  XX'  and  YY'  occur,  they  are  fixed  straight 
lines  intersecting  at  right  angles  at  0,  the  letters  being  arranged  as  in  the 
figure  of  §  270  ;  that  all  positive  or  negative  real  numbers  ar3  repre- 
sented by  lines  to  the  right  or  left  of  O,  respectively,  in  XX' ;  and  all 
positive  or  negative  pure  imaginary  numbers  by  lines  above  or  below  0, 
respectively,  in  YY'. 

429.   Graphical  Representation  of  Complex  Numbers. 

We  will  now  show  how  to  repre- 
sent the  complex  number  a  +  hi. 

Let  a  be  represented  by  OA,  to  the 
right  of  0  if  a  is  positive,  to  the  left 
if  a  is  negative. 

Let  hi  be  represented  by  OB,  above 
0  if  6  is  positive,  below  if  h  is 
negative.  ^^ 

Draw^  line  AC  equal  and  parallel  to  OB,  on  the  same  side  of 
XX'  as  OB,  and  line  OG. 

Then,  OG  is  considered  the  result  of  adding  hi  to  a ;  that  is, 
OG  represents  a  +  hi. 


B 


C'^"" 


y^ 


\)V 


0    a    A 


Y' 


256  ADVANCED  COURSE  IN  ALGEBRA 

This  agrees  with  the  methods  already  given  for  adding  a  real  or  pure 
imaginary  number,  if  either  a  or  6  is  zero. 

The  figure  represents  the*  case  where  a  and  h  are  both  positive  ;  if  a  is 
positive  and  6  negative,  OC  will  lie  between  OX  and  OF ;  if  a  is  nega- 
tive, 00  will  lie  between  OY  and  OX'  if  h  is  positive,  and  between  OX^ 
and  Oy  if  6  is  negative. 

In  accordance  with  §  §  57  and  406,  —  (a  +  hi)  may  be  repre- 
sented by  line  0C\  where  OO  is  equal  in  length  to  OC,  and 
drawn  in  the  opposite  direction  from  0. 

430.  The  modulus  of  a  real,  pure  imaginary,  or  complex 
number  is  the  length  of  the  line  which  represents  the  number. 

The  argument  is  the  angle  between  the  line  which  represents 
the  number  and  OX,  measured  from  OX  in  a  direction  opposite 
to  the  motion  of  the  hands  of  a  clock. 

If,  ^  for  example,  in  the  figure  of  §  429,  ZXOC  =  30%  the 
argument  of  the  complex  number  represented  by  OG  is  30°, 
and  the  argument  of  the  complex  number  represented  by  OC 
is  210°. 

The  modulus  is  always  taken  positive,  and  the  argument 
may  have  any  value  from  0°  to  360°. 

The  real  numbers  +  a  and  —  a  have  the  modulus  a,  and  arguments  0° 
and  180°,  respectively ;  the  pure  imaginary  numbers  +  ai  and  —  ai  have 
the  modulus  a,  and  arguments  90°  and  270°,  respectively. 

431.  In  the  figure  of  §  428,  00  =  -^ OA^  +  AG^  =  Vo^+T"; 
that  is,  the  modulus  of  the  complex  number  a-{-biis  Va^  +  6^; 
this  is  also  the  modulus  of  the  complex  numbers  a— hi,  —a-\-bi, 
and  —a  —  hi. 

432.  Graphical  Representation  of  Addition. 

We  will  now  show  how  to  represent  the  result  of  adding  h 
to  a,  where  a  and  h  are  any  two  real,  pure  imaginary,  or  com- 
plex numbers. 

Let  a  be  represented  by  OA,  and  h  by  OB.  /         ~~/!^ 

Draw  line  AG  equal  and  parallel  to  OB,  /  ^y^ I 

on  the  same  side  of  0^1  as  OB,  and  line  OG.         /  y^    / 

Then,  OG  is  considered  the  result  of  add-       //         / 
ing  h  to  a\  that  is,  line  OG  represents  a-\-h.    o      a     ji 


PURE  IMAGINARY  AND  COMPLEX  NUMBERS     257 


This  agrees  with  the  method  of  §  429,  which  is  a  special  case  of  the 
above. 

In  like  manner,  the  sum  of  any  number  of  real,  pure  imagi- 
ary,  or  complex  numbers  may  be  represented  by  a  straight 
line  drawn  from  0. 

433.  Graphical  Representation  of  Subtraction. 
Let  a  and  h  be  any  two  real,  pure  imagi- 
nary, or  complex  numbers. 

Let  a  be  represented  by  OA^  and  h  by 
OB ;  and  complete  parallelogram  OB  AG. 

By  §  432,  OA  represents  the  result  of 
adding  the  number  represented  by  OB  to 
the  number  represented  by  OC. 

That  is,  if  h  be  added  to  the  number 
represented  by  OC,  the  sum  is  equal  to 
a ;  hence,  a  —  6  is  represented  by  line  OC. 

434.  Graphical  Representation  of  Multiplication. 

Since  +  ai  may  be  written  (  +  1)  x  {-\-ai),  the  product  of  +1 
by  -\-  ai  is  represented  by  turning  the  line 
OA,  which  represents  the  number  + 1, 
through  one  right  angle,  in  a  direction 
opposite  to  the  motion  of  the  hands  of  a 
clock,  and  multiplying  the  result  by  a. 

And  since  —ai  may  be  written  (+1)  x 
(—  ai),  the  product  of  +1  by  —  ai  is  repre- 
sented by  a  line  equal  in  length  to  that 
which  represents  the  product  of  + 1  by 
+  aiVbut  drawn  in  the  opposite  direction  from  0. 

This  suggests  the  following : 

The  product  of  any  real,  pure  imaginary,  or  complex  number 
by  +  ai  may  be  represented  by  turning  the  line  which  repre- 
sents the  number  through  one  right  angle,  in  a  direction  oppo- 
site to  the  motion  of  the  hands  of  a  clock,  and  multiplying  the 
result  by  a. 


Y 

B- 


+1^ 

-ai 
■B' 


258 


ADVANCED  COURSE  IN   ALGEBRA 


c'a^ 


The  product  of  any  real,  pure  imaginary,  or  complex  number 
by  —  at  may  be  represented  by  a  line  equal  in  length  to  the 
line  which  represents  its  product  by  +  at,  but  drawn  in  the 
opposite  direction  from  0. 

435.  Since  a  +  bi  may  be  written  (+  1)  x  (a  +  6i),  the  prod- 
uct of  -f  1  by  a  +  bi  is  represented  by 

turning  the  line  OA,  which  represents 
the  number  + 1,  through  an  angle 
equal  to  the  argument  of  a-\-bi  (§  430), 
in  a  direction  opposite  to  the  motion 
of  the  hands  of  a  clock,  and  multi- 
plying the  result  by  the  modulus  of 
a  -\-  bi. 

And  since  —  (a  -f  bi)  may  be  written  (+1)  x  (—a—  bi),  the 
product  of  + 1  by  —  (a  -f-  bi)  is  represented  by  a  line  equal  in 
length  to  that  which  represents  the  product  of  +1  by  a  +  bi, 
but  drawn  in  the  opposite  direction  from  0. 

This  suggests  the  following : 

If  a  and  b  are  any  real  numbers,  the  product  of  any  real, 
pure  imaginary,  or  complex  number  by  a  -f  bi  may  be  repre- 
sented by  turning  the  line  which  represents  the  number  through 
an  angle  equal  to  the  amplitude  of  a  -f  bi,  in  a  direction  oppo- 
site to  the  motion  of  the  hands  of  a  clock,  and  multiplying 
the  result  by  the  modulus  of  a  -h  bi. 

The  product  of  any  real,  pure  imaginary,  or  complex  number 
by  —  (a  -f-  bi)  may  be  represented  by  a  line  equal  in  length  to 
the  line  which  represents  its  product  by  a  -f-  bi,  but  drawn  in 
the  opposite  direction  from  0. 

436.  Let  a  and  b  be  any  two  real,  pure 
imaginary,  or  complex  numbers,  repre- 
sented by  the  lines  OA  and  OB,  respec- 
tively. 

The  result  of  multiplying  a  by  6  is 
represented  by  line  OC,  where  angle 
XOC  is  the  sum  of  angles  XOA  and 
XOB,  and  OC  =  OA  x  OB  (§  435). 


PURE   IMAGINARY  AND   COMPLEX  NUMBERS     259 


Z  XOA  -  Z  XOB,  and  OC  ■ 


That  is,  ab  is  represented  by  OC. 

In  like  manner,  the  product  of  any  number  of  real,  pure 
imaginary,  or  complex  numbers  may  be  represented  by  a 
straight  line  drawn  from  0. 

437.  Graphical  Representation  of  Division. 

Let  a  and  b  be  any  two  real,  pure  imaginary,  or  complex 
numbers. 

Let  a  be  represented  by  OA,  and  b 
by  OB. 

Draw  line  OC,  making  ZXOC  = 

OA 
OB 

Then,  Z  XOA  ==  Z  XOC  +  Z  XOB, 
and  OA=OCx  OB. 

Whence,  by  §  436,  OA  represents  the  product  of  the  number 
represented  by  OC  by  the  number  represented  by  OB. 

Then,  OC  represents  a  number  which,  when  multiplied  by  b, 

gives  a;  and  hence  0(7  represents  -" 

b 
Therefore,  the  quotient  of  any  two  real,  pure  imaginary,  or 
complex  numbers  can  be  represented  by  a  straight  line  drawn 
from,  0. 

438.  Graphical  Representation  of  Roots. 

Let  a  be  any  real,  pure  imaginar}^,  or  complex  number, 
represented  by  line  OA. 

Draw  line  OB  making  Z  XOB  =  ~Z  XOA, 

n 

and  having  its  length  equal  to  the  ?ith  root 
of  the  modulus  of  a. 

Then,  Z  XOA  =  nxZ  XOB,  and  the 
modulus  of  a  is  the  7ith  power  of  the  length 
of  OB. 

Then,  by  §  436,  OA  represents  the  nt\\  power  of  the  number 
represented  by  OB. 

Whence,  OB  represents  Va.  ' 


260  ADVANCED   COURSE   IN   ALGEBRA 

439.  It  follows  from  §§  432,  433,  436,  437,  and  438  that  any 
number  which  is  the  result  of  any  finite  number  of  the  follow- 
ing operations  performed  upon  one  or  more  real,  pure  imaginary, 
or  complex  numbers,  may  be  represented  by  a  straight  line 
drawn  from  0 1 

Addition;  Subtraction;  Multiplication;  Division ^  raising  to 
any  power  whose  exponent  is  a  rational  number  (§  51) ;  extract- 
ing any  root. 

This  is  a  graphical  representation  of  the  fact  that  any  such 
number  can  be  expressed  in  the  form  a  -j-  6^,  where  a  and  b 
are  real  numbers,  either  of  which  may  be  zero.  (Compare 
§§417  and  804.) 

We  shall  limit  ourselves  in  the  present  work  to  numbers  of  the  above 
form. 

The  discussion  of  complex  exponents  requires  a  knowledge  of  Higher 
Trigonometry. 


QUADRATIC  EQUATIONS  261 


XIX.    QUADRATIC  EQUATIONS 

440.  A  Quadratic  Equation  is  an  equation  of  the  second  degree 
(§  113),  with  one  or  more  unknown  numbers. 

In  the  present  chapter  we  consider  only  quadratic  equations 
involving  one  unknown  number. 

The  principles  demonstrated  in  §§  116  to  119,  inclusive,  and  §  122,  hold 
for  quadratic  equations. 

441.  By  transposing  all  terms  to  the  first  member,  any 
quadratic  equation,  involving  one  unknown  number,  oj,  may  be 
reduced  to  the  form     „^  +  j^  +  ,  ^  0. 

If  neither  b  nor  c  is  zero,  this  is  called  a  Complete  Quadratic 
Equation. 

A  Complete  Quadratic  Equation  is  sometimes  called  an  Affected  Quad- 
ratic Equation. 

If  either  or  both  of  the  numbers  b  and  c  are  zero,  the  equa- 
tion is  called  an  Licomplete  Quadratic  Equation. 

An  incomplete  quadratic  equation  of  the  form  ax^  -f-  c  =  0,  is 
called  a  Pure  Quadratic  Equation. 

In  §  183,  we  showed  how  to  solve  quadratic  equations  of  the  forms 

aa;2  _^  5^;  =  0,  ax^  +  c  =  0,  a;2  +  ax  +  6  =  0,  and  ax'^-\-hx  +  c  =  0, 
when  the  first  members  could  be  resolved  into  factors. 

442.  Consider  the  equation 

where  ^  is  a  rational  and  integral  expression  involving  the 
unknown  numbers. 

We  may  write  the  equation, 

A^-&  =  0,  or  {A  +  B){A-B)  =  0. 

By  §  182,  the  latter  is  equivalent  to  the  set  of  equations 

^  +  ^  =  0,  and  ^-jB  =  0. 


262      ADVANCED  COURSE  IN  ALGEBRA 

Or,  to  the  equations  A=:-\-B,  and  A  =  —  B,  which  may  be 
written  together  in  the  form 

A  =  ±B. 

The  sign  ± ,  called  the  double  sign,  when  prefixed  to  a  number,  indi- 
cates that  it  may  be  either  +  or  — . 

Thus,  the  given  equation  is  equivalent  to  an  equation  which 
is  obtained  by  equating  the  positive  square  root  of  the  first 
member  to  ±  the  square  root  of  the  second. 

PURE  QUADRATIC  EQUATIONS 

443.  A  pure  quadratic  equation  may  be  solved  by  reducing 
it,  if  necessary,  to  the  form  x^=a,  and  then  equating  ic  to  ±  Va 

(§  442). 

1.  Solve  the  equation  3x^ -\-7  =  —  +  35. 

4 

Clearing  of  fractions,  12  a;^  +  28  =  5  a;^  + 140.        \ 

Transposing  and  uniting  terms,  7  x'  =  112.  X 

Dividing  by  7,  x-  =  16. 

Equating  ic  to  ±  the  square  root  of  16,  x=±  4. 

2.  Solve  the  equation  7 x'^  —  5  =  5xF  —  13. 

Transposing  and  uniting  terms,  2  a;^  =  —  8. 

Or,  a;2  =  -  4. 

Equating  a;  to  ±  the  square  root  of  —4,  a;=±V— 4 

=  ±2V=T(§409). 

In  this  case,  both  values  of  x  are  imaginary  (§  407) ;  it  is  impossible  to 
find  a  real  value  of  x  which  will  satisfy  the  given  equation. 

EXERCISE  49 

Solve  the  following : 

1.   2(:l  x  -  5)2 +  3(a;  + 10)2  =  434.        2.    (a;  +  1)8 -(x  -  1)8  =  20. 

3     §^'  +  —  +  —  =  ^^^-  —  -5 
*      «         10       12        12  8 

4.  (2  X  +  7 )  (o  .X  -  6)  -  (4  a;  -  3)  (7  a;  +  5)  -  24  X  +  59  =  0. 

5.  (x  +  2  a)  (a;  +  3  6)  +  (X  -  2  a)  (a;  -  3  6)  =  a;2  +  4  a2  +  9  62. 


QUADRATIC   EQUATIONS  263 

6    3  a;2  +  7      5  a;2  +  3      4  cc^  _  ip  ^  ^ 
7  14  35  ' 

7.  (X  +l){x-  2)(x  -  3)-(3c  -  l){x  +  2)(x  +  3)  =  -52. 

COMPLETE  QUADRATIC  EQUATIONS 

444.   First  Method  of  Completing  the  Square. 

By  transposing  the  terms  involving  x  to  the  first  member, 
and  all  other  terms  to  the  second,  and  then  dividing  both  mem- 
bers by  the  coefficient  of  x^,  any  complete  quadratic  equation 
can  be  reduced  to  the  form 

Q?  -\-  px  =  q. 

A  trinomial  is  a  perfect  square  when  its  first  and  third  terms 
are  perfect  squares  and  positive,  and  its  second  term  plus  or 
minus  twice  the  product  of  their  square  roots  (§  167). 

Then,  the  square  root  of  the  third  term  is  equal  to  the  second 
term  divided  by  twice  the  square  root  of  the  first. 

Hence,  the  square  root  of  the  expression  which  must  be  added 

to  x^  -\-px  to  make  it  a  perfect  square,  is  ^,  or  ^, 
Adding  to  both  members  the  square  of  ^,  we  have 

^'  +  P^  +  f  =  ^  +  f  =  ^^^-T-^-     , 

By  §  442,  this  is  equivalent  to  an  equation  which  is  obtained 
by  equating  the  positive  square  root  of  the  first  member  to  ± 
the  square  root  of  the  second. 

Then,  it  is  equivalent  to 


Adding  to  both  members  such  an  expression  as  will  make  the  first 
member  a  perfect  trinomial  square,  is  called  Completing  the  Square. 

445.  We  derive  from  §  444  the  following  rule  for  solving  a 
complete  quadratic  equation. 


264  ADVANCED  COURSE  IN  ALGEBRA 

Reduce  the  equation  to  the  form  x^  -\-px  =  q. 

Complete  the  square,  by  adding  to  both  members  the  square  of 
one-half  the  coefficient  of  x. 

Equate  the  positive  square  root  of  the  first  member  to  ±  the 
square  root  of  the  second,  and  solve  the  linear  equations  thus 
formed. 

446.   1.   Solve  the  equation  3cc^  —  8a;  =  —  4. 

Dividing  by  3,  a:2_^  =  _|. 

Adding  to  both  members  the  square  of  -,  we  have 

^_8£+m^  =  »4     16^1 
S       [SJ  3      9      9 

Equating  the  positive  square  root  of  the  first  member  to  ± 
the  square  root  of  -,  4.         o 

3         3 
Transposing  - 1,  ic  =  |  ±  |  =  2  or  |. 

If  the  coefiicient  of  x^  is  negative,  the  sign  of  each  term  must 
be  changed. 

2.   Solve  the  equation    —  9  o;^  —  21  a;  =  10. 
Dividing  by  -  9,  aj2  +  Z^  =  _^. 

7  \ 

Adding  to  both  members  the  square  of  -,  \ 

3       [fij  9      36     36 

Equating  the  positive  square  root  of  the  first  member  to  ± 
the  square  root  of  — >  /  rr         o 

wi,  7,325 

Whence,  ^  ""  ~  6  "^6^  ~3  °''  ~  3* 


QUADRATIC   EQUATIONS  265 

447.  By  §  167,  aV-f-  6aj  will  become  a  perfect  trinomial  square 

by  adding  to  it  the  square  of  r-^,  or  - — 

2  ax         2  a 

Hence,  we  may  complete  the  square  by  adding  to  both  mem- 
bers the  square  of  the  quotient  obtained  by  dividing  the  coefficient 
of  X  by  twice  the  square  root  of  the  coefficient  ofx^. 

This  is  usually  a  more  convenient  rule  than  that  of  §  445, 
when  the  coefficient  of  x^  is  a  perfect  square. 

1.   Solve  the  equation  9  a^  —  5  ic  =  4. 

5 


Adding  to  both  members  the  square  of 


2x3' 


\^J  36      36 

5  13 
Extracting  square  roots,  3  a;  —  -  =  ±  — -. 

6  6 

Then,  3:K  =  |±^  =  3'or-|. 

Dividing  by  3,  ic  =  1  or .' 

y 

If  the  coefficient  of  a?  is  not  a^erfect  square,  it  may  be 
made  so  by  multiplication. 

2.   Solve  the  equation  8^  — 15  a;  =  2. 
Multiplying  each  term  by  2,  16  a^  —  30  a;  =  4. 
Adding  to  both  members  the  square  of ,  or  — , 


16a^-30a.  +  ^_j=4  +  _  =  _. 


15  17 

Extracting  square  roots,  4  a; =  ±  — . 

4  4 

Then,  4aj  =  l^±lI  =  8or-i. 

4       4  2 


266  ADVANCED   COURSE   IN   ALGEBRA 

Whence,  ic  =  2  or  — . 

'  8 

If  the  coefficient  of  x^  is  negative,  the  sign  of  each  term  must  be 
changed. 

448.   Second  Method  of  completing  the  Square. 

Every  complete  quadratic  equation  can  be  reduced  to  the 
form  aa^  +  6x  +  c  =  0,  or  ax- -\-bx=  —c  (^  441). 
Multiplying  both  members  by  4  a,  we  have 

4  aV  +  4  abx  =  —  4  ac. 


Completing   the   square   by  adding  to   both   members   the 

-  (§  447),  or  b,  we  obtain 
I 

4  aV  +  4  abx  +  6^  =  ft^  —  4  a/i. 


square  of  (§  447),  or  b,  we  obtain 

2  X  2  a 


Extracting  square  roots,  2  ax-\-b=  ±  V  6^  —  4  ac. 


Transposing,  2ax=  —b  ±  V 6^  —  4  ac. 

-6±  V6'-4a.c 


Whence,  x  = 


2a 


We  derive  from  the  above  the  following  rule  for  completing 
the  square: 

Reduce  the  equation  to  the  form  ax^  +  6x  +  c  =  0. 
Multiply  both  members  by  four  times  the  coefficient  of  x^,  and 
add  to  each  the  square  of  the  coefficient  of  x  in  the  given  equation. 

The  advantage  of  this   method   over   the   preceding  is  in 
avoiding  fractions  in  completing  the  square. 

449.   1.    Solve  the  equation  2x^  —  lx=  —  3. 
Multiplying  both  members  by  4  x  2,  or  8, 

Wx'-mx=-24t. 
Adding  to  both  members  the  square  of  7, 

16  a^-  56  a;  +  7^=  -  24  +  49  =  25. 


QUADRATIC   EQUATIONS  267 

Extracting  square  roots,  4  a?  —  7  =  ±5. 

4  a;  =  7  ±  5  =  12  or  2. 

Whence,  ■        c»  =  3  or  ^. 

If  the  coefficient  of  x  in  the  given  equation  is  even^  fractions 
may  be  avoided,  and  the  rule  modified,  as  follows : 

Multiply  both  members  by  the  coefficient  ofx^,  and  add  to  each 
the  square  of  half  the  coefficient  of  x  in  the  given  equation. 

2.   Solve  the  equation  15  «2  +  28  a;  =  32. 
Multiplying  both  members  by  15,  and  adding  to  each  the 
square  of  14,  we  have 

152a;2  + 15(28  x)  +  14^  =  480  + 196  =  676. 
Extracting  square  roots,   15  a;  +  14  =  ±  26. 

15a;=-14±26=12or  -40. 

Whence,  a;  =  -  or  — -• 

The  method  of  completing  the  square  exemplified  in  the  present  section 
is  called  the  Hindoo  Method. 

EXERCISE  50 

The  following  may  be  solved  by  either  of  the  preceding  methods, 
preference  being  given  to  the  one  best  adapted  to  the  example  under 
consideration. 

1.  x2-a;  =  12.  8.  4  a;2  -  12  x  =  23. 

2.  3a;2-17a;  =  -10.  9.  8a:2  +  7x  +  2  =  0, 

3.  5«2  +  i7x  =  12.  10.  36x2  +  3x  =  6. 

4.  10x2  + 27a; +  14  =  0.  11.  28x2  +  29x  +  6  =  0. 

5.  36x2-24x  =  77.  12.  10  -  37a;  =  - 30x2. 

6.  49a;2  +  21a;-4  =  0.  13.  a;(5x  +  22)+ 36  =(2x  + 5)2. 

7.  6  +  23x-18x2  =  0.  14.  (2x  +  l)8-(2x+3)3=_386. 
16.    (3a;  +  2)(2x-3)  =  (4x-l)2-14. 

16.  (a;  +  4)(2x-l)  +  (2x-l)(3x  +  2)-(3x  +  2)(4x-l)  =  -49. 

17.  (x  +  1)  (a:  +  3)  =  (x  +  7)  V2  +  12. 

18.  (5  +  2V3)x2-(4  +  14V3)x  =  9- 14V3.'' 


268  ADVANCED   COURSE   IN   ALGEBRA 

450.   Solution  of  Complete  Quadratic  Equations  by  Formula. 

It  follows  from  §  448  that,  if 

ax^  -f-  &a^  4-  c  =  0, 

then  ^^-6±Vy-4ac,  ^^^ 

Z  CI 

This  result  may  be  used  as  a  formula  for  the  solution  of  any 
complete  quadratic  equation  in  the  form  ao?  +  6aj  +  c  =  0. 

1.  Solve  the  equation  2a;^  +  5  a;  —  18  =  0. 

Here,  a  =  2,  6  =  5,  and  c  =  —  18 ;  substituting  these  values 
in  (1),  we  have 

^^-5±V25Tll^-5±13^^^^_9, 

4  4  2  I 

2.  Solve  the  equation  110  a;^  —  21  a;  =  —  1. 
Here,  a  =  110,  6  =  -21,  c  =  1. 

Then,      ^^21±VSr3440^21±2^1_^^l, 
'  220  220        10        11 

Dividing  both  terms  of  the  fraction  in  equation  (1)  by  2, 


-  =  -^-^ (§377)=^-Ai (2) 


This  is  a  convenient  formula  in  case  the  coefficient  of  x  in 
the  given  equation  is  even. 

3.   Solve  the  equation  —  5  o;^  + 14  a?  +  3  =  0. 
Here,  a  =  —  5,  6  =  14,  c  =  3 ;  substituting  in  (2), 

—  5  —  5  5 

Particular  attention  must  be  paid  to  the  signs  of  the  coefl&cients  in 
making  the  substitution. 

The  student  should  now  work  the  examples  of  Exercise  50 
by  formula. 


I 


QUADRATIC  EQUATIONS  269 

451.   Fractional  Equations  involving  Quadratics. 

In  solving  fractional  equations  which  involve  quadratics,  we 
reject  any  solution  which  satisfies  the  equation  obtained  by 
equating  to  zero  the  L.  C.  M.  of  the  given  denominators  (§  222). 

3  2 

1.   Solve  the  equation =  1. 

x—6     x—o 

Multiplying  each  term  by  (x  —  6)(x  —  5),  we  have 

3a;-15-2a;  +  12  =  a;2_i;I^^_l_3Q 

Or,  a^-12x  =  -S3. 

Completing  the  square,    aj^  — 12  a;  +  6^  =  3. 

Extracting  square  roots,  x  —  6  =  ±  V3. 

Whence,  x=6±  V3. 

Since  neither  6  +  V3  nor  6  —  V3  satisfies  the  equation 
(x  —  6)  (cc  —  5)  =  0,  both  roots  may  be  retained. 

1  1         a^^Sx-6 


2.    Solve  the  equation 
We  may  write  it 


2^x     2-x  a^-4 

1       ,      1         a^  +  3a;-6 


x-\-2     x-2  a^-4 

Multiplying  each  term  by  a;^  —  4, 

x-2  +  x  +  2  =  x^-\-Sx-6. 
Or,  x^  +  x==6. 

Multiplying  by  4,  and  adding  V  to  both  members, 

4a;2  +  4x  +  l  =  25. 

Extracting  square  roots,  2  a;  + 1  =  ±  5. 

Then,  2x  =  -l±5  =  4or  -6. 

Whence,  a;  =  2  or  —  3. 

Since  2  satisfies  the  equation  a^  —  4  =  0,  it  must  be  rejected, 
and  the  only  solution  is  x  =  —  3. 

452.   Literal  Equations  involving  Quadratics. 

In  solving  literal  equations  which  involve  complete  quad- 
ratics, the  methods  of  §  449  are  usually  the  most  convenient. 


270  ADVANCED  COURSE   IN   ALGEBRA 

Ex.    Solve  the  equation  acx^  —  adx  -\-  hex  —bd  =  0. 

We  write  the  equation  acxr  —  (ad  —  bc)x  =  bd. 

Multiplying  by  4  ac,  and  adding  (ad  —  bey  to  both  members, 

4  a^c^x^  —  4  ac  (ad  —  bc)x-\-  (ad  —  bey 

=  4  abed  +  a'd'  -  2  abed  +  6V  =  a'd'+  2  abed  +  bh\ 
Extracting  square  roots, 
2  acx  —  (ad  —  be)  =  ±  (ad  +  be). 

2 aca;  =  ad—be±  (ad -{-be)  =2 ad  or  —  2  be. 

Whence,  a;  =  -or 

c  a 

If  several  terms  contain  the  same  power  of  x,  the  coefficient  of  that 

power  should  be  enclosed  in  parentheses,  as  shown  above. 


The  above  equation  may  be  solved  more  easily  by  the  method  of  §  183 ; 
thus,  by  §  156,  the  equation  may  be  written 

(ax  +  b)  (ex  —  d)=0. 

Then,  ax  +  b  =  0,  or  a:  =  —  - ; 

a 

and  ex  —  d  =  0,  or  a;  =  — 

c 

Several  equations  in  Exercise  51  may  be  solved  most  easily  by  the 

method  of  §  183.  

As  a  general  rule,  literal  quadratics  are  best  solved  by  formula. 

453.  Equations  leading  to  Quadratics,  having  the  Unknown 
Number  under  Radical  Signs. 

In  solving  equations  of  this  kind,  only  solutions  which 
satisfy  the  given  equation  should  be  retained. 

In  verifying  solutions,  only  principal  values  of  the  roots^are 
considered  (§  397). 

-t  n 

1.    Solve  the  equation  VS  -\-x  +  VS  —  x  =  —         • 

Vs  —  x 

Clearing  of  fractions,      ■^^25  —  x^  +  5  —  cc  =  12. 


Transposing,  V25  —  x-  =  x-{-7. 

Squaring,  25  —  x^  =  x'^ -}- 14:  x -\-  49. 


QUADRATIC   EQUATIONS  271 

Or,  2a^-l-14a;=:-24. 

Multiplying  by  2,  and  adding  7^  to  both  members, 

4x2  _^  28x4-49  =  1. 
Extracting  square  roots,         2  x  +  7  =  ±  1. 

2x  =  -7±l  =  -6  or  -8. 
Whence,  a;  =  —  3  or  —  4. 

Putting  X  =  —  3,  the  given  first  member  becomes 

V2  +  V8  =  V2  +  2  V2  =  3  \/2. 

12      3x2x2  /~ 

The  second  member  becomes  — ::  = ' = —  =  3  v  2. 

V8  2\/2 

Then,  the  solution  x  =  —  3  is  correct. 

Putting  X  =  —  4,  the  given  first  member  becomes  1  +  \/9  =  4. 

12 
The  second  member  becomes  —  =  4. 

V9 
Then,  the  solution  x  =  —  4  is  correct. 


2.   Solve  the  equation  Vx  —  1  +  V3  x  +  3  =  4. 
Transposing  -y/x  —  1,       V3  x  +  3  =  4  —  Vx  —  1. 


Squaring,  S  x-^3  =  W-SWx-l  +  x-l. 

Transposing,  8  Vx  —  1  =  12  —  2  x. 

Or,  4  Vx  — 1  =  6  —  X. 

Squaring,  ■l6a;-16  =  36-12x  +  a^. 

Or,  0^-28  x  =  -52. 

Completing  the  square, 
^  ic2-28x  +  142  =  144. 

Extracting  square  roots,      aj  — 14  =  ±  12. 
Whence,  x  =  14  ±  12  =  26  or  2. 

Putting  X  =  26,  the  given  first  member  becomes  5  +  9,  or  14. 
Then,  the  solution  x  =  26  is  not  correct. 
Putting  X  =  2,  the  given  first  member  becomes  1  +  3,  or  4. 
Then,  x  =  2  is  the  only  correct  solution. 


272  ADVANCED   COURSE   IN   ALGEBRA 


EXERCISE  51 

Solve  the  following : 

1.  2x_A=:I^_2i.  14.   V5x2-3x-41  =  3a;-7. 

3       4x      9       4a: 


5        13       1 


15.  6  -  V5  a;2  _  9  =  12. 


6^     9x2     18  16.   V7  a;  +  8  -  V5  x  -  4  =  2. 

3.  A  +  i^  =  _§^.  17  _A«___^+AL.  =  o 

4x3  18  *a;-6&3a-10& 

4.  x2  +  2  wa;  =  1  -  m2.  ^g       1  1     _      a;2  -  17 


5.  a;2-2ax  =  -6a  +  9.  ^  +  3     a;-5     a;2-2x-15 

6.  a;2  +  ax  +  &x  =  -  ab.  19.      ^  ^  +  ^^"^^  ^   =  _-l^ 

7.  cc2  —  (w  —  l)a:  =  7i. 


a;  -  2     24  (x  +  2)      x^  -  4 
x-2 _  x  +  4 _  _  7 

8.  6x  +  5_4x  +  4^Q  *";x+5     x-3~     3* 

4x-3         X-3  '  o        r     o 

n-i    X  —  a     x  +  a_x2  —  5a2 

9.  x2  —  m2wx  +  mn^x  =  m^n^. 


10.  x2  -  4  ax  -  10  X  =  -  40  a. 
^^    10a;2-3^6x2  +  6     Gx^-l 


a     X  —  a      x2  —  .a2 

22.   V(a  +  2  6)x-2a6  =  x  -  4  6. 
21 


18  9  9  x2  -  2'      23.  2  X  +  \/4  x2  -  7 


V4  x2  -  7 

12.  6  X?  +  4  ax  -  15  6x  =  10  a6.  „  „  . 

24,        ^ "        _  *_ 

13.  aTOx2  +  anx  +  bmx  +  bn  =  0.  '2x  +  a     3x  —  4a     3 

25.  2  \/3  X  +  4  +  3  V3  X  +  7  =        ^ 

26.  (a  +  x)3  +  (6-x)3  =  (a  +  6)3.       32. 


V3x  +  4 
1  1  14 


x2  _  3  X      x2  +  4  X      15  x2 

27.  V^'=r^  +  V2x  +  3a  =  \/6a.         33.  2x_±i  +  §A=_2  ^ IZ . 

3x-2     2x+l.     4 

28.  3Vx^-— i=_  =  4.  34    __J^  =  l_l4.1. 

Vx  —  1  a  —  6  +  x     a     b     x 

29.  _^  +  «±^  =  2(«!±&!l.  35    If-i ^Usf-i 1^ 

a-\-b         X  a'^-b''  3V4x-l     2/        VSx+l      3^ 

30   x^-Jl^_4a6_.  3g   ^^ ^-^  =  1  +     ' 


X         a2-62  aj2-4     3(x  +  2)  2-x 

31.  J^ iziA^  =  _5  37.  3V^T^  +  2V^T6^  =  -24a. 


4-5x        3x  6 

3  x2  -  4     4  x2  +  3         9  x2  -  123 


x2  +  6       2  x2  -  1      2  X*  +  9  x2  -  5 


=  1. 


QUADRATIC   EQUATIONS 


273 


5       .       7       ^  8  x^  -  13  a;  -  64 
*    2a;  +  3'^3x-4        6x''  +  x-12 


40. 


x  +  4a       Zx  —  2a 


30  ax  —  (jp- 


2x-Sa      3x  +  «       6a;2-7ax-3a2 


41. 


X  -{-  a 


x+  b  —  c 


=  2. 


45. 


x  +  4 


5+  4x 


x-1      x2  +  x  +  l      x^ 


2. 


42.   V^  +  V(m-n)x  +  mn  =  2m.   45     3x^  +  x-2  ^3x^  +  4x-l^ 

2x^-4x-6     2x2-2x-l 
47.  aV(l  +  x)2  -  62<^2(i  _  ^y  =  0. 
2x+l       2n+l 


43    28(3x+10)  25 

'      8  x3  -  27         2  x2  -  3  X " 
2  X  -  3  ?i       3x  +  n  _  10_ 
3x  +  7i       2x-3?i      3 
1 


44. 


48 


50. 


Vl  +  x2  +  Vl 

x         X  +  4  ^ 
x  +  4 


\/x  +  1      Vn  +  1 

1 ^  V2, 

X2* 


\/r+x2- VI 


X 


51.   V6  -  5  X  +  V2  -  7  X  =  V12  +  6  x. 

5.  ^+_«  +  ^Zl«  =  «+_^  +  ^Jl^.      63.   V¥T^-^/^T^  =  a-b. 
X  —  a     x-\-  a     a  —  h     a  —  b 

54.  (X  -  2  a  +  3  6)2  +  (X  +  3  a  -  2  6)2  =  25  a^  -  30  a6  +  13  62. 

55.  VxMn  +  V^^l  _  V^MH  -  Vg^  ^  2  V16^^^. 


Vx2  +  1 


1     \/x2+T  4-  Vx2  -  1 


56.  (a  -  6  +  2  c)x2  -  (2  a  +  6  +  c)x  =  -  a  -  2  6  +  c. 


57.  Va2  +  ax  +  x2  +  Va2  -  ax  +  x2  =  a  ( V7  +  Vl3). 


x+i     x^h^     x  +  3^3^ 
a;_l^a:-2     x-3 


59   a:  +  5     x-7     a;  +  9_.^^ 
■x-4     x+6     x-8 


60.  Vl  +  X  +  x2  +  Vl  -  X  +  x2  =  X  (1  +  V3). 

61.  (X  4-  a  -  2  6)3  -  (X  -  2  a  +  6)3  =  9  (a  -  6)3. 
««       6  1  2  3 


x+3     x-5     x+7 


9 


63.  V3  x2  +  4  X  +  10  -  V3  x2  +  2  X  -  8  =  2. 


64. 


-^  +  l  +  -i-  +  l  =  o, 

x  +  a     a     x  +  6     6 


65. 


+u 


1 


0. 


X     6  —  X     a     a  +  6 
66.  (3  mn  +  n"^) x^-(Gm^-l  iim  -  2  n2)x  +'2  win  -  3  n2  =  0. 


67. 


1      x+3     x+2     x+4 


x+7     x+5     x+8     x+6 


274  ADVANCED   COURSE  IN  ALGEBRA 

68.   (8  a'' -j- 10  ab-S  62)  x^  -  (16  a"- -^  6  b"^)  x  +  S  a^  -I0ab-Sb^  =  0. 

69        ^     4-      ^     —  ^  _L  5 .  70    ^  ~  ^  _|_  ^  —  ^  _  X  —  4:     a;  —  5 

x—a     x—b     b     a  'x—2     x—S     x—5     x—6 


71.   Va:2  +  2  X  -  Sr  -  Vx^  -  2  x  -  3  =  V2  x^  -  6. 


.  72.   \/x2  +  7  ax  +  12  a^  +  Vx'^  -  7  ax  +  12  a'^  =  V2  x^  +  24  a^. 

73. ^ + ^ + ^ + ^ =  0. 

x+2a+36     x-2a-3&     x+2a-3  6     x-2a+36 

^^    x'^  +  6  X  +  1      4  x^  4- 12  X  +  1  _  2  x^  -  12  X  -  1      3  x-  -  9  x  -  1 . 
x+6  x+3  x-6  x-3 

76.  (x-i]  (x-?)  ^x-^]  =  (X  +  2)(x  +  3)(x  +  4). 

77.  (x2  +  l)(a2  -  &2  _  c2  +  2  &c)  =  2 x(a2  +  62  _  2 6c  +  c2). 


THEORY   OF  QUADRATIC  EQUATIONS 

454.   A  quadratic  equation  cannot  have  more  than  two  different 
roots. 

Every  quadratic  equation  can  be  reduced  to  the  form 

ax^  4-  6ic  +  c  =  0. 

If  possible,  let  this  equation  have  three  different  roots,  rj,  rg, 
and  ^3 ;  then,  by  §  110, 

ar^'  +  hr^  +  c  =  0,  (1) 

ar/-}-6r2  +  c  =  0,  (2) 

and  ar^  +  6?  3  +  c  =  0.  (3) 

Subtracting  (2)  from  (1), 

Then,         a{r^  +  r^  {r^  -r^  +  h  (n  -  r^)  =  0, 

or,  (ri  —  rg)  {ar^  +  arg  +  6)  =  0. 

Whence,  either  r^  —  7-2  =  0,  or  else  ar^  +  a?-2  +  &  =  0  (§  49). 
But  rj  —  rs  cannot  be  zero,  for,  by  hypothesis,  r^  and  rg  are 
different. 

Hence,  ar^  -{-  ar^  +  h  =  0,  (4) 


QUADRATIC  EQUATIONS  276 

In  like  manner,  by  subtracting  (3)  from  (1),  we  have 

m\  4-  arg  +  &  =  0.  (5) 

Subtracting  (5)  from  (4),  arg  —  ar^  =  0,  or  rg  — Tg  =  0. 

But  this  is  impossible,  for,  by  hypothesis,  rg  and  r^  are 
different. 

Hence,  a  quadratic  equation  cannot  have  more  than  two 
different  roots. 

455.   Sum  and  Product  of  Roots. 

Let  Ti  and  7*2  denote  the  roots  of  ax^  -\-bx-\-  c  =  0. 
Then,  by  §  450, 


2a  '  '  2a 

Adding  these  values,    ri  +  Tz  =  ^ —  = 

2a  a 

Multiplying  them  together, 

4ta^  4a^      a 

Hence,  if  a  quadratic  equation  is  in  the  form  ax^  -\-bx-\-c  =  0, 
the  sum  of  the  roots  equals  minus  the  coefficient  of  x  divided 
by  the  coefficient  of  x^,  and  the  ^woduct  of  the  roots  equals  the 
independent  term  divided  by  the  coefficient  ofx^. 

1.  Find  by  inspection  the  sum  and  product  of  the  roots  of 

3a.'2_7a;-15  =  0. 

7  —15 

The  sum  of  the  roots  is  -,  and  their  product  — -— ,  or  —  5. 
o  o 

17 

2.  One  root  of  the  equation  6  a;^  +  31  a;  =  —  35  is  —  -  ;  find 

the  other. 

The  equation  can  be  written  6  or'  +  31  a;  +  35  =  0. 

q-J  OK 

Then,  the  sum  of  the  roots  is ,  and  their  product  — 

Then,  the  other  root  is  _  §1  + 1,  or  —  -^  -  - :  that  is,  -  -• 

6      2        6  2  3 


276  ADVANCED   COURSE   IN  ALGEBRA 


EXERCISE  52 

Find  by  inspection  the  sum  and  product  of  the  roots  of : 

1.  a:2  4-  7  a;  +  6  =  0.  5.  4  x  -  12  a;^  =  3. 

2.  x2  -  jc  +  12  =  0.  6.  9  X  -  21  x2  +  7  =  0. 

3.  2c2  +  3  a:  -  1  =  0.  7.  6  a;2  +  x  +  4  =  0. 

4.  3a;2-a;-6  =  0  8.   6  ax'^  +  7  abx  =  2Q  b\ 
9.  One  root  of  x2  +  7  x  =  98  is  7  ;  find  the  other. 

10.  One  root  oi28x'^  -  x  -  15  =  0  is  --  ;  find  the  other. 

11.  One  root  of  x'^-9x  +  xVS  +  2-7V3  =  0is5  +  2V3;  find  tlie 
other. 

12.  If  ri  and  ^2  are  the  roots  of  ax'^  +  bx  +  c  =  0,  express  the  following 
in  terms  of  ri  and  rz  : 

(a)  ri2  +  rir2  +  r^^.       (&)  ni±l2! .       (c)    i^  +  i^ .        (c?)  n^  +  r^z, 

nr^  ri2     ^22 

456.   Discussion  of  General  Solution. 

By  §  448,  the  roots  oi  ax^ -\- bx -\- c  =  0  are 


2a  2a 

We  will  now  discuss  these  results  for  all  possible  real  values 
of  a,  6,  and  c. 

I.  6^  —  4  ac  positive. 

In  this  case,  r^  and  rg  are  rea?  and  unequal. 

II.  6^  _4ac  =  0. 

In  this  case,  r^  and  r2  are  real  and  egi^a?. 

III.  hi  — 4,  ac  negative. 

In  this  case,  r^  and  rg  are  imaginary  (§  407),  or  complex 
(§  415). 

IV.  6  =  0. 

In  this  case,  the  equation  takes  the  form 

ax^  +  c  =  0 ;  whence,  x=  ±\\  —  -» 

^      a 

If  a  and  c  are  of  unlike  sign,  the  roots  are  real^  equQil  irk 

absolute  valuer  and  unlike  in  sign. 


QUADRATIC   EQUATIONS  277 

If  a  and  c  are  of  like  sign,  both  roots  are  imaginary, 

V.  c  =  0. 

In  this  case,  the  equation  takes  the  form 

ax^  +  bx  =  0:  whence,  x  =  0,  or 

a 

Hence,  the  roots  are  both  reol^  one  being  zero. 

VI.  6  =  0,  a7idc  =  0. 

In  this  case,  the  equation  takes  the  form  aoiy^  =  0. 
Hence,  both  roots  equal  zero. 

The  roots  are  both  rational,  or  both  irrational,  according  as 
6^  —  4  ac  is^  or  is  not,  a  perfect  square. 

Ex.     Determine  by  inspection  the  nature  of  the  roots  of 

2x2-505-18  =  0. 
Herea  =  2,6  =  -5,c  =  -18;  and  6^  -  4  ac  =  25 -}- 144  =  169. 
Since  b^  —  Aac  is  positive,  the  roots  are  real  and  unequal. 
Since  6^  —  4  ac  is  a  perfect  square,  both  roots  are  rational. 

EXERCISE  53 

Determine  by  inspection  the  nature  of  the  roots  of : 

1.  6  x^  +  7  a;  -  5  =  0.  6.    16  x2  -  9  =  0. 

2.  4x2 -x  =  0.  7.   9x2  =  12x  +  l. 

3.  10a;2  +  i7a;  +  3  =  0.  8.   25x2  +  30x  +  9  =  0. 

4.  4x2 -20  a; +  25  =  0.  9.  7  a;2  +  3  a;  =  0. 

6.   a;2  -  21  x  + 200  =  0.  10.   41  x- 20x2  =  20. 

FACTORING 

457.   Factoring  of  Quadratic  Expressions. 
A  quadratic  expression  is  an  expression  of  the  form 
ax^  +  hx-\-c. 

In  §  174,  we  showed  how  to  factor  certain  expressions  of  this 
form  hy  inspection;  we  will  now  derive  a  rule  for  factoring  any 
quadratic  expression. 


278  ADVANCED   COURSE   IN   ALGEBRA 

We  have, 

ax^  -\-bx  -^  c  =  afx^  +  —  -^  -\ 
\         a      aj 

L         a      \2aJ      4:0?     aJ 

by  §171.     (1) 
But  by  §  450,  the  roots  oi  ax^ -{-hx-\-c  =  0  are      ^ 

h       ^b^-4:ac  ^^^         b       Vb^-4:ac 
2a  2a  2a  2a        ' 

Hence,  to  factor  a  quadratic  expression,  place  it  equal  to  zero, 
and  solve  the  equation  thus  formed. 

Theii  the  required  factors  are  the  coefficient  of  x^  in  the  given 
expression,  x  minus  the  first  root,  and  x  minus  the  second. 

1.   Factor  Q>x^  +  1  x-Z. 

Solving  the  equation  6  a;^  +  7  a;  —  3  =  0,  by  §  450, 


-7±V49  +  72_-7±ll     1^,      3 

''-  12 --"l2~"==3  °'  "2* 

Then,  ^x^j^7x-S  =  q[x-^(x-\-^ 

=  (3a;-l)(2a;  +  3). 
The  example  may  also  be  solved  by  using  (1)  as  2l  formula. 

2.    Factor  4  +  13  a;  -  12  a^. 

Solving  the  equation  4  +  13  a;  —  12  aj^  =  0,  we  have 


^_  -13±V169  +  192^-13±19^     14 
-24  -24  43' 


QUADRATIC   EQUATIONS  279 

Whence,    4:-\-lSx-12x'=z-12fx-^^fx-^ 

-<-i)K-')e-i) 

=  (1  +  4  a;)  (4  -  3  x). 

3.   ¥2iGtov2x^-Sxy-2y^-7x  +  4.y  +  6. 
We  solve  the  equation 

2x^-x{3y-h7)-2y^  +  4:y-\-6  =  0. 


By  §450,      ^^3y  +  7±V(3y  +  7)^  +  16/-32y-48 

^3y  +  7±V25y^  +  10y  +  l 
4 

^3y  +  7±(5y  +  l) 
4 

=    -^  ^     or  ^-J—  =  2  ?/  -f  2  or  — ^-^ — 

4  4  -^  2 

Then,         2x''-3xy-2y^-7x  +  4.y-{-6 

=  2[x-(2y  +  2)-]^x-^y^'] 

=  (x-2y-2){2x-\-y-3). 

458.  If  the  coefficient  of  x^  is  a  perfect  square,  it  is  prefer- 
able to  factor  the  expression  by  completing  the  square  as  in 
§  447,  and  then  using  §  171. 

1.   Factor  9  a^- 9  a; -4. 

By  §  447,  9  ic^  —  9  a?  will  become  a  perfect  square  by  adding 

9  3 

to  it  the  square  of  - — -,  or  -;  then, 
2x3         2 

9a52_9a._4  =  9a^_-9a;+/'|Y-?-4 

-i''-2)-r 


280  ADVANCED   COURSE  IN  ALGEBRA 

Then,      9x'-9x-4.==(3x-^  +  ^Vsx-^-^\    (§171) 

=  {3x-\-l){3x-i). 

If  the  x"^  term  is  negative,  the  entire  expression  should  be 
enclosed  in  parentheses  preceded  by  a  —  sign. 

2.  Factor  3 -12  a;- 4  ccl 

3  _  12  a;  -  4  a;2  =  -  (4  a;2  4- 1 2  a;  -  3) 

=  _(4a;2  +  l2a;  +  9-9-3) 

=  -[(2a;  +  3)2-12] 

=  (2a;  +  3+Vl2)  X  (-l)(2iv  +  3-VT2) 

=  (2V3  +  3  +  2  a;)  (2 V3  -  3  -  2  x). 

In  certain  cases,  the  factors  of  a  quadratic  expression  involve 
complex  numbers. 

3.  Factor  a;2  -  4  a;  +  9.  ^ 

x'-4,x-\-9  =  (x'-Ax  +  4)  +  5 

=  (x-2f-(-^)  _ 

=  (a;  _  2  +  V- 5)  (a;  -  2  -  V- 5). 

EXERCISE  54 

Factor  the  following  by  the  method  of  §  457 : 

1.  a;2+14a;  +  33.  10.    8  3:24.18  a; -5. 

2.  a;2  -  13  ic  +  40.  11.    6  a;2  +  7  x  +  2. 

3.  x^-x-  42.  12.    12  a;2  +  7  X  -  45. 

4.  4  a;2  +  3  X  -  7.  13.    14  x2  -  23  xa  +  3  a^. 

5.  3x2-llx-20.  14.   24  x2  -  17  xy  +  3  ?/2. 

6.  2  x2  +  9  X  +  9.  15.   28  x2  -  mx  -  2  m^. 

7.  5x2-36x  +  36.  16.   5  -  26  x- 24x2. 

8.  9-8x-x2.  17.   8  +  14 X- 15x2. 

9.  20  +  19x-6x2.  18.   21x2  +  23x?/2  +  6y*. 

19.   x^-xp-^6y'^-6x  +  lSy  +  5. 


QUADKATIC   EQUATIONS  281 

20.  x^-Sxy-iy^  +  6x-iy  +  S. 

21.  x^-6xy-\-5y^-2x-2y-S. 

22.  2  a2  +  5  a&  +  2  &2  +  7  a  +  5  6  +  3. 

23.  Zx^-\-7xy-6y^-10xz-8yz  +  S  z\ 

24.  2-7?/-7x  +  3z/2  +  x^-4x2. 

Factor  the  following  by  the  method  of  §  468  : 

25.  4  a;2  -  12  a;  -  7.  30.  1  +  2  a;  -  %\ 

26.  32  -  12  a;  -  9  x2.  31.  16  ^2  -  16  x  +  1. 

27.  16  x2  +  56  a:  +  33.  32,  25  a;2  -  25  ic  +  6. 

28.  9  x2  +  24  X  -  2.  33.  36  ^2  +  72  x  +  29. 

29.  4x2  +  20x  +  19.  34.  11  +  10  x- 25x2. 

459.  We  will  now  take  up  the  factoring  of  expressions  of 
the  forms  a?*  +  aa^?/^  +  2/^  or  a;*  +  2/^  when  the  factors  involve 
surds.     (Compare  §  172.) 

1.  Factor  a^  +  2  a^^^  +  25  &^ 

a*  +  2  a^h^  +  25  6^  =  {p>  +  10  a%^  +  25  6^)  -  8  o?}}" 

=  (a2  +  5  62)2_(a5V8)' 

.  =  (a^  +  5  62  +  a6 V8)(a2  +  5  6^  _  a6V8) 

=  (a2  +  2  a6  V2  +  5  62)(a2  -  2  ab-^2  +  5  62). 

The  above  expression  may  also  be  expressed  as  the  product  of  two 
factors  involving  complex  numbers. 

2.  Factor  x'' ^\. 

a^  + 1  =  (iB^  +  2a^  + 1)  -  2  a^ 
=  (aj2  +  l)2_(a;V2)2 
=  (a;2  _j_  3,^2  4-  V){y?  -  ajV2  + 1). 

EXERCISE  55 

In  each  of  the  following,  obtain  two  sets  of  factors,  where  this  can  be 
done  without  bringing  in  imaginary  numbers  : 

1.  x4-7x2  +  4.  4.   4a4  +  6a2  +  9. 

2.  otS  +  64.  5.  36  x*  -  92  x2  +  49. 

3.  9  m*-  11  w2  +  1.  6.   26  m*  +  28  m2n2  +  16  w*. 


282       ADVANCED  COURSE  IN  ALGEBRA 

460.   Solution  of  Equations  by  Factoring. 

In  §  183,  we  showed  how  to  solve  equations  whose  first  mem- 
bers could  be  resolved  by  inspection  into  first  degree  factors, 
and  whose  second  members  were  zero. 

We  will  now  take  up  equations  whose  first  members  can  be 
resolved  into  factors  partly  of  the  first  and  partly  of  the  second, 
or  entirely  of  the  second  degree. 

1.    Solve  the  equation  ic^  —  1  =  0. 
Factoring  the  first  member,  {x  —  l){x^  -\-  x  -{-1)  :=  0. 
Then,  a;  —  1  =  0,  or  a;  =  1 ; 

and  a:-  +  a;  +  1  =  0 ;  whence,  by  §  450, 


l±Vl-4      -1±V- 


2  2 

2.    Solve  the  equation  a;''  +  1  =  0. 
By  Ex.  2,  §  459, 

a;4  -f  1  =  {f  -f  a; V2  -\-l){x' -  x^2  +  1). 
Solving    x^  +  a; V2  +  1  =  0,  we  have 


-V2±V2^^      -V2±V-2 


2 

Solving    a?  —  a; V2  +  1  =  0,  we  have 


^^V2±V2-4^V2±V-2^ 


EXERCISE  56 


Solve  the  following  equations  : 

1.  3  a:3- 2x2 +15  a; -10  =  0.  2.  x*  +  4a;2  -  32  =  0. 

3.  x3  -  64  =  0.  5.    27  x3  +  8  =  0.  7.  64  x^  -  125  =  0. 

4.  3  a;4  +  24  X  =  0.           6.    16  x*  -  81  =  0.            8.  x^  -  729  a^  =  0. 
g  a;2  +  2x  +  4^4_             ^^    x\x  +  4  a)  -  9  g^  x^  -  2  g^  ^  ^ 

'  x2  -  2  X  -  4     x^'  *   x2(x  -  4  a)  -  9  a3  a:2  +  2  a^ 

11.  x4  +  81=0.  13.    x*  + 2x2 +  25  =  0.     15.  x^  -  256  «»  =  0. 

12.  x4  -  5  x2  +  1  =  0.       14.    x4  -  18  x2  +  9  =  0.     16.  9  x*  -  x2  +  4  =  0. 


r  QUADRATIC   EQUATIONS  283 

461.   Maxima  and  Minima  Values  of  Quadratic  Expressions. 

The  greatest  or  least  value  of  a  quadratic  expression  may 
sometimes  be  found  by  the  artifice  of  completing  the  square. 

1.    Find  the  minimum  value  oi  x^  —  5x-\-7. 

We  have  x^  —  5x-j-7  =  (x ] -\ — 

Since  (x J  is  positive  for  every  real  value  of  x,  the  least 

value  of    x  —  -]  +  -  is  when  x  = 


2;    '  4  2  3 

Thus,  the  minimum  value  of  the  expression  is  -• 

2.    Find  the  maximum  value  of  4  —  3  ic  —  2  a^. 
We  may  write  the  expression 


4^2la^-\-lx^  =  4.-2[fx-{-^^'      ^ 


2     1  I  V        47      16 

3 


41      of    .  3 

¥-T  +  4 


The  greatest  value  is  when  x  = 

.    41 

Thus,  the  greatest  value  of  the  expression  is  —  • 

8 

EXERCISE  57 

Find  the  maxima  and  minima  values  of  the  following,  and  determine 
which : 

1.  x'^  +  Sx-l.  3.   4x2-8x-5.  5.   Sx^  +  5x  +  4. 

2.  6-Sx-x^.  4.    3  +  a;-x2.  6.    -  2  -  9a: -9x2. 

7.   6x2 -7  a; +  3.  8.    -7  +  2x-5x2. 

PROBLEMS  INVOLVING  QUADRATIC   EQUATIONS   WITH 
ONE   UNKNOWN  NUMBER 

462.  In  solving  problems  which  involve  quadratic  equations, 
there  will  usually  be  two  values  of  the  unknown  number ;  only 
those  values  should  be  retained  which  satisfy  the  conditions  of 
the  problem. 

The  considerations  of  §§  261  and  262  hold  for  equations  of 
any  degree. 


284      ADVANCED  COURSE  IN  ALGEBRA 

1.  A  man  sold  a  watch  for  $21,  and  lost  as  many  per  cent 
as  the  watch  cost  dollars.     Find  the  cost  of  the  watch. 

Let  X  =  number  of  dollars  the  watch  cost. 

Then,  x  =  the  per  cent  of  loss, 

X  x^ 

and  X  X  — ,  or =  number  of  dollars  lost. 

100'        100 

By  the  conditions,  —  =  a;  —  21. 

100 

Solving,  aj  =  30  or  70. 

Then,  the  cost  of  the  watch  was  either  $30  or  $70  ;  for  either  of  these 
answers  satisfies  the  conditions  of  the  problem. 

2.  A  farmer  bought  some  sheep  for  f  72.  If  he  had  bought 
6  more  for  the  same  money,  they  would  have  cost  him  $1 
apiece  less.     How  many  did  he  buy  ? 

X  =  number  bought.  ■    . 

number  of  dollars  paid  for  one. 


Let 

X 

Then, 
and 

72 

X 

72 
cc  +  6 

By  the  conditions. 

72 

X 

Solving, 

X 

=  number  of  dollars  paid  for  one  if  there 
had  been  6  more. 

x  +  6 
=  18  or  -  24. 

Only  the  positive  value  of  x  is  admissible,  for  the  negative  value  does 
not  satisfy  the  conditions  of  the  problem. 

Therefore,  the  number  of  sheep  was  18. 

If,  in  the  enunciation  of  the  problem,  the  words  "6  more"  had  been 
changed  to  "6  fewer,"  and  "^1  apiece  less"  to  "$1  apiece  more," 
we  should  have  found  the  answer  24.     (Compare  §  261.) 

EXERCISE  58 

1.  What  number  added  to  its  reciprocal  gives  2^  ? 

2.  Divide  the  number  24  into  two  parts  such  that  twice  the  square  of 
the  greater  shall  exceed  5  times  the  square  of  the  less  by  45. 

3.  Find  three  consecutive  numbers  such  that  the  sum  of  their  squares 
shall  be  434. 

4.  Find  two  numbers  whose  difference  is  7,  and  the  difference  of 
whose  cubes  is  1267. 


QUADRATIC  EQUATIONS  285 

5.  rind  five  consecutive  numbers  such  that  the  quotient  of  the  first 

by  the  second,  added  to  the  quotient  of  the  fifth  by  the  fourth,  shall 

23 

equal  —  • 

^        12 

6.  Find  four  consecutive  numbers  such  that  if  the  sum  of  the  squares 
of  the  second  and  fourth  be  divided  by  the  sum  of  the  squares  of  the  first 

13 

and  third,  the  quotient  shall  be  — • 

7.  The  area  of  a  certain  square  field  exceeds  that  of  another  square 
field  by  1008  square  yards.  And  the  perimeter  of  the  greater  exceeds  one- 
half  that  of  the  smaller  by  120  yards.     Find  the  dimensions  of  each  field. 

8.  A  fast  train  runs  8  miles  an  hour  faster  than  a  slow  train,  and  takes 
3  hours  less  to  travel  288  miles.    Find  the  rates  of  the  trains. 

9.  The  perimeter  of  a  rectangular  field  is  184  feet,  and  its  area  1920 
square  feet.     Find  its  dimensions. 

10.  A  merchant  sold  goods  for  $  22.75,  and  lost  as  many  per  cent  as 
the  goods  cost  dollars.     What  was  the  cost  ? 

11.  A  merchant  sold  two  pieces  of  cloth  of  different  quality  for  $  40.25, 
the  poorer  containing  28  yards.  He  received  for  the  finer  as  many  dollars 
a  yard  as  there  were  yards  in  the  piece ;  and  7  yards  of  the  poorer  sold 
for  as  much  as  2  yards  of  the  finer.    Find  the  value  of  each  piece. 

12.  A  merchant  sold  goods  for  $50.69,  and  gained  as  many  per  cent 
as  the  goods  cost  dollars.     What  was  the  cost  ? 

13.  A  has  five-fourths  as  much  money  as  B.  After  giving  A  $  6,  B's 
money  is  equal  to  A's  multiplied  by  a  fraction  whose  numerator  is  15, 
and  whose  denominator  is  the  number  of  dollars  A  had  at  first.  How 
much  had  each  at  first  ? 

14.  A  and  B  set  out  at  the  same  time  from  places  247  miles  apart, 
and  travel  towards  each  other.  A  travels  at  the  rate  of  9  miles  an  hour  ; 
and  B's  rate  in  miles  an  hour  is  less  by  3  than  the  number  of  hours  at  the 
end  of  which  they  meet.     Find  B's  rate. 

15.  A  man  buys  a  certain  number  of  shares  of  stock,  paying  for  each 
as  many  dollars  as  he  buys  shares.  After  the  price  has  advanced  as  many 
dimes  per  share  as  he  has  shares,  he  sells,  and  gains  $  722.50.  How  many 
shares  did  he  buy  ? 

16.  The  two  digits  of  a  number  differ  by  1  ;  and  if  the  square  of  the 
number  be  added  to  the  square  of  the  given  number  with  its  digits  reversed, 
the  sum  is  585.     Find  the  number. 

17.  A  gives  $  336,  in  equal  amounts,  to  a  certain  number  of  persons. 
B  gives  the  same  sum,  in  equal  amounts,  to  18  fewer  persons,  and  gives 
to  each  $  6  more  than  A.     How  much  does  A  give  to  each  person  ? 


286  ADVANCED   COURSE  IN  ALGEBRA 

18.  The  telegraph  poles  along  a  certain  road  are  at  equal  intervals. 
If  the  interval  between  the  poles  were  increased  by  22  feet,  there  would 
be  8  fewer  in  a  mile.     How  many  are  there  in  a  mile  ? 

19.  A  merchant  bought  a  cask  of  wine  for  1 45.  Having  lost  3  gallons 
by  leakage,  he  sells  the  remainder  at  f  1.50  a  gallon  above  cost,  and  makes 
a  profit  of  33^  per  cent  on  his  entire  outlay.  How  many  gallons  did  the 
cask  contain  ? 

20.  The  men  in  a  regiment  can  be  arranged  in  a  column  twice  as  long 
as  it  is  wide.  If  their  number  were  224  less,  they  could  be  arranged  in  a 
hollow  square  4  deep,  having  in  each  outer  side  of  the  square  as  many  men 
as  there  were  in  the  length  of  the  column.     Find  the  number  of  men. 

21.  The  denominator  of  a  fraction  exceeds  twice  the  numerator  by  2, 
and  the  difference  between  the  fraction  and  its  reciprocal  is  — •  Eind 
the  fraction.  ^"^ 

22.  A  man  started  to  walk  3  miles,  intending  to  arrive  at  a  certain 
time.  After  walking  a  mile,  he  was  detained  10  minutes,  and  was  in 
consequence  obliged  to  walk  the  rest  of  the  way  a  mile  an  hour  faster. 
What  was  his  original  speed  ? 

23.  A  regiment,  in  solid  square,  has  24  fewer  men  in  front  than  when 
in  a  hollow  square  6  deep.     How  many  men  are  there  in  the  regiment  ? 

24.  A  rectangular  field  is  surrounded  by  a  fence  160  feet  long.  The 
cost  of  this  fence,  at  96  cents  a  foot,  was  one-tenth  as  many  dollars  as 
there  are  square  feet  in  the  area  of  the  field.  What  are  the  dimensions 
of  the  field? 

25.  A  crew  can  row  down  stream  18  miles,  and  back  again,  in  7|  hours. 
Their  rate  up  stream  is  1|  miles  an  hour  less  than  the  rate  of  the  stream. 
Find  the  rate  of  the  stream,  and  of  the  crew  in  still  water. 

26.  A  man  put  $  5000  in  a  savings-bank  paying  a  certain  rate  of  in- 
terest. At  the  end  of  a  year  he  withdrew  1 75,  leaving  the  remainder  at 
interest.  At  the  end  of  another  year,  the  amount  due  him  was  f  5278.50. 
Find  the  rate  of  interest. 

27.  A  square  garden  has  a  square  plot  of  grass  at  its  centre,  surrounded 
by  a  path  4  feet  in  width.  The  area  of  the  garden  outside  the  path  ex- 
ceeds by  768  square  feet  the  area  of  the  path ;  and  the  side  of  the  garden 
is  less  by  16  feet  than  three  times  the  side  of  the  plot,  Find  the  dimen- 
sions of  the  garden. 

28.  A  merchant  has  a  cask  full  of  wine.  He  draws  out  6  gallons, 
and  fills  the  cask  with  water.  Again  he  draws  out  6  gallons,  and  fills 
the  cask  with  water.  There  are  now  25  gallons  of  pure  wine  in  the  cask. 
How  many  gallons  does  the  cask  hold  ? 


QUADRATIC   EQUATIONS  287 

29.  A  and  B  sell  a  quantity  of  corn  for  ^  22,  A  selling  10  bushels  more 
than  B.  If  A  had  sold  as  many  bushels  as  B  did,  he  would  have  received 
$  8  ;  while  if  B  had  sold  as  many  bushels  as  A  did,  he  would  have  received 
$  15.     How  many  bushels  did  each  sell,  and  at  what  price  ? 

30.  Two  men  are  employed  to  do  a  certain  piece  of  work.  The  first 
receives  $48  ;  and  the  second,  who  works  6  days  less,  receives  $27.  If 
the  second  had  worked  all  the  time,  and  the  first  6  days  less,  they  would 
have  received  equal  amounts.  How  many  days  did  each  work,  and  at 
what  wages  ? 

31.  A  and  B  run  around  a  course,  starting  from  the  same  point,  in 
opposite  directions.  A  reaches  the  starting-point  4  minutes,  and  B  9 
minutes,  after  they  have  met  on  the  road.  If  they  continue  to  run  at 
the  same  rates,  in  how  many  minutes  will  they  meet  at  the  starting- 
point  ? 

32.  A  carriage- wheel,  15  feet  in  circumference,  revolves  in  a  certain 
number  of  seconds.  If  it  revolved  in  a  time  longer  by  one  second,  the 
carriage  would  travel  14400  feet  less  in  an  hour.  In  how  many  seconds 
does  it  revolve  ? 

DISCUSSION    OF    PROBLEMS    INVOLVING    QUADRATIC 
EQUATIONS   WITH  ONE  UNKNOWN  NUMBER 

463.  Interpretation  of  Complex  Results. 

Prob.  Let  it  be  required  to  find  two  real  numbers  whose 
sum  shall  be  10,  and  product  26. 

Let  X  =  one  number. 

Then,  10  -  x  =  the  other. 

By  the  conditions,         a;  (10  -  a;)  =  26. 

Solving,  x  =  5  ±  V—  1. 

"We  conclude  that  the  given  conditions  cannot  be  satisfied,  and  the 
problem  is  impossible. 

Hence,  imaginary  or  complex  results  show  that  the  problem  is 
impossible. 

464.  The  Problem  of  the  Lights. 

To  find  upon  the  line  which  joins  two  lights,  A  and  B,  the 
point  equally  illuminated  by  them;  it  being  given  that  the 
intensity  of  a  light,  at  a  certain  distance,  equals  its  intensity 
at  the  distance  1,  divided  by  the  square  of  the  given  distance. 


288  ADVANCED  COURSE  IN  ALGEBRA 

1 \ \ 1 [ 

C"  A  OB  C 

Let  J5  be  c  units  to  the  right  of  A. 

Let  'a  and  h  denote  the  intensities  of  A  and  J5,  respectively, 
at  the  distance  1. 

Let  the  point  of  equal  illumination  be  x  units  to  the  right 
oiA. 

Then  it  will  be  c  —  a;  units  from  B. 

By  the  conditions  of  the  problem,  the  intensity  of  A  at  the 

distance  x  units,  is  —;  and  the  intensity  of  B  at  the  distance 

•*/ 

c  —  X  units,  is 


Then, 


(c  -  xy 

a 


^      (c  -  xf 


c  Va        ^^        c  Va 


Solving  this  equation,   x  =  — ::   or 


Va  -I-  Vft  Va  —  V6 

Since  there  are  two  lights,  c  must  always  be  positive ;  then, 
neither  a,  6,  nor  c  can  equal  zero. 

The  problem  then  admits  of  only  three  different  hypotheses : 

1.  a  >  6. 

In  this  case,  ^^-^  is  <  1,  and  >  -• 

Va4-V6  2 

Then,  the  first  value  of  aj  is  <  c,  and  >  |- 

Thus,  the  first  point  of  equal  illumination  is  at  (7,  between 
the  lights,  and  nearer  5,  the  lesser  light. 

Again,      ^    ^        is  >  1. 

Va  —  V6 
Then,  the  second  value  of  cc  is  >  c. 

Thus,  the  second  point  of  equal  illumination  is  at  C",  in  AB 
produced,  to  the  right  of  the  lesser  light. 

2.  a<h. 

In  this  case,  — ^^ — -  is  <  -,  and  the  first  value  of  x  <  ^. 
Va  4-  V6  ^  ^ 


QUADRATIC   EQUATIONS 


289 


Then,  the  first  point  of  equal  illumination  is  between  the 
lights,  and  nearer  A,  the  lesser  light. 

Again,  the  second  value  of  x  is  negative. 

Then,  the  second  point  of  equal  illumination  is  at  O",  in  BA 
produced,  to  the  left  of  the  lesser  light. 

3.   a  =  b. 
In  this  case, —  =  -,  and  the  first  value  of  a?  =  -. 

Then,  the  first  point  of  equal  illumination  is  midway  between 
the  lights. 

c-y/a 


Again, 


-  =  00  (§247). 


Va—  V& 

Then,  there  is  no  second  point  of  equal  illumination  in  AlBj 
or  AB  produced. 

In  this  case,  as  Va  —  Vb  approaches  the  limit  zero,  the  second  value 
of  X  increases  without  limit. 

That  is,  as  the  difference  between  the  intensities  of  the  lights  approaches 
the  limit  zero,  the  distance  from  A  to  the  second  point  of  equal  illumina- 
tion increases  without  limit. 


GRAPHICAL    REPRESENTATION    OF    QUADRATIC    EXPRES- 
SIONS WITH  ONE  UNKNOWN  NUMBER 

465.   The  graph  of  a  quadratic  expression,  with  one  unknown 
number,  may  be  found  as  in  §  279. 
Ex.   Eind  the  graph  of  ar^  —  2  a;  —  3. 
Put  2/ =  a^- 2a; -3. 

Ifa;=0,       y  =  -S.    (A) 

Ifa;  =  l,       2/  =  -4.    (B) 

Ifa;  =  2,      y  =  -3.    (O) 

Ifa;  =  3,      y  =  0.        (D) 

lix  =  4:,       y  =  5.       (E) 

ltx  =  -l,  y  =  0.        (F) 

ltx  =  -2,  y  =  5.        (G)  , 
The  graph  is  the  curve  QBE. 


290  ADVANCED  COURSE  IN   ALGEBRA 

By  taking  other  values  for  x,  the  curve  may  be  traced  beyond 
E  and  G. 

It  extends  in  either  direction  to  an  indefinitely  great  dis- 
tance from  XX'. 

To  determine  the  lowest  point  of  the  curve,  we  must  know 
what  negative  value  of  y  has  the  greatest  absolute  value ;  this 
may  be  found  as  in  §  461. 

We  have,  a^  _  2  ic  -  3  =  (a;  - 1)^  -  4. 

The  latter  expression  has  its  negative  value  of  greatest  abso- 
lute value  when  x  =  l,  being  then  equal  to  —  4. 

Then,  the  lowest  point  of  the  curve  has  the  co-ordinates 
(1,  —  4)  ;  and  is  therefore  the  point  B. 

466.  The  principle  of  §  280  holds  for  the  graph  of  the  first 
member  of  any  quadratic  equation,  with  one  unknown  number. 

Thus,  the  above  graph  intersects  XX'  twice ;  once  at  x  =  3, 
and  once  at  aj=  —  1 ;  and  the  roots  of  the  equation x^—2x—3  =  0 
are  3  and  —  1. 

467.  Graphs  of  the  First  Members  of  Quadratic  Equations 
having  Equal  or  Complex  Roots. 

1 .   Consider  the  equation  ic^ — 4  a; + 4 = 0. 
By  §  183,  the  two  roots  are  2  and  2. 
To  find  the  graph  of  the  first  member, 
put  y  =  (x—2y. 

If  a;  =  0,  2/  =  4.     It  x  =  2,  y  =  0.  x'- 

If  a;  =  1,  2/  =  1.     If  »  =  3,  2/  =  1 ;  etc. 

The  graph  is  the  curve  ABC,  which 
extends   beyond  A  and   G  to  an   indefinitely  great  distance 
from  XX'. 

Since  y  cannot  be  negative,  the  graph  is  tangent  to  XX'  at 
the  point  whose  co-ordinates  are  (2,  0). 

It  is  evident  from  this  that,  if  a  quadratic  equation,  with 
one  unknown  number,  has  equal  roots,  the  graph  of  its  first 
member  is  tangent  to  XX'. 


QUADRATIC   EQUATIONS 


291 


2.    Consider  the  equation  a^-fa^-f  2  =  0. 


Solving,     X ; 

To  find  the  graph  of  the  first  member, 

put  2/  =  a;^  +  a;  +  2. 

If  a;  =  0,  2/  =  2.     If  a;  =  -  1,  2/  =  2. 

If  aj  =  1,  2/  =  4.     If  a;  =  —  2,  2/=  4;  etc. 

The  graph  is  the  curve  ABC,  which 
extends  beyond  A  and   C  to  an   indefinitely  great  distance 
from  XX'. 

To  find  the  point  B,  where  the  curve  is  nearest  to  XX',  we 
have  /       1\2     7 

The  latter  expression  has  its  least  value,  ^,  when  x  =  —  -' 

4  Z 


Then,  B  has  the  co-ordinates 


i-iiy 


It  is  evident  from  this  that,  if  a  quadratic  equation,  with 
one  unknown  number,  has  complex  roots,  the  graph  of  its  first 
member  does  not  intersect  XX'. 

The  equation  a'^x^  —  b'^  =  0  may  be  written  (ax  +  b)(ax  —  b)  =  0. 
The  graph  of  the  first  member  in  this  case  is  a  pair  of  straight  lines 

parallel  to  YY',  respectively,  -  to  the  right  and  -  to  the  left,  of  that 
line. 

EXERCISE  59 

.    Find  the  graph  of  the  first  member  of  each  of  the  following,  determine 
its  lowest  point,  and  verify  the  principles  of  §§  280  and  467  in  the  results  : 

1.  a:-2-5a;  +  4  =  0.     3.    4a:2  +  7x  =  0.  5.   4x2  +  12x  +  9  =  0. 

2.  x'^  +  x-6  =  0.       4.    8x2- 14x- 15=0.     6.    2x2  +  4x  +  5  =  0. 

Find  the  graphs  of  the  first  members  of  the  following,  and  verify  the 
principle  of  §  280  in  the  results  : 

7.   x2-16  =  0.  8.   9x2-25  =  0. 


292  ADVANCED  COUKSE  IN  ALGEBRA 


XX.    EQUATIONS  SOLVED  LIKE  QUAD- 
RATICS 

468.   Equations  in  the  Quadratic  Form. 

An  equation  is  said  to  be  in  the  quadratic  form  when  it  is  in 

where  n  is  an  integer  or  fraction ;  as, 

and  x-^  +  x~^  =  72. 

Such,  equations  are  readily  solved  by  the  methods  of  the  pre- 
ceding chapter. 

1.  Solve  the  equation  a?^  —  6  a?^  =  16. 
Completing  the  square  by  the  rule  of  §  445, 

Extracting  square  roots,        a^  —  3  =  ±  5. 

Then,  a^  =  3±5  =  8or  -2. 

Extracting  cube  roots,    .  a;  =  2  or  —  V2. 

There  are  also  four  imaginary  roots,  which  may  be  found  by  the  method 
of  §  460. 

2.  Solve  the  equation  2  a;  +  3  Vaj  =  27. 

Since  -Vx  is  the  same  as  x^,  this  is  in  the  quadratic  form. 
Multiplying  by  8,  and  adding  3^  to  both  members, 

16  a;  +  24  V^  +  9  =  216  +  9  =  225. 

Extracting  square  roots,  4  Vaj  +  3  =  ±  15. 

Then,  4V^  =  -3±15  =  12or -18. 


Or,  •  ya^  =  3or-?. 

But  only  principal  roots  are  considered. 


EQUATIONS  SOLVED  LIKE   QUADRATICS         293 

Therefore,  -\Jx  cannot  be  negative,  and  the  only  solution  is 

Vx  =  3,  or  a;  =  9. 
In  solving  an  equation  of  the  form 

flKK*  +  6a;«+  c  =  0, 
p 
any  value  of  aj«  which  is  not  a  principal  root,  should  be  rejected. 

_8 

For  example,   if  a;  ^  =  —  2,  the  corresponding  solution   should   be 
rejected. 

3.    Solve  the  equation  2  x'^  - 11  x'^  + 12  =  0. 
By  formula  (1),  §  450, 

^-|_11±V121~96^11±5^^       3. 
4  4  2 

In  this  case,  neither  value  is  rejected. 

Extracting  square  roots,  x^  =  ±  2  or  ±  (  - )  • 
Eaising  t»  third  power,  a;~^  =  ±  8  or  ±  ( -V* 

Inverting,  .       a;  =  ±  -  or  ±  /^-V  • 

p 
To  solve  an  equation  of  the  form  xi  =  a,  first  extract  the  root  corre- 
sponding to  the  numerator  of  the  fractional  exponent,  and  afterwards 
raise  to  the  power  corresponding  to  the  denominator ;  careful  attenti<m 
must  be  given  to  algebraic  signs. 

469.   An  equation  may  sometimes  be  solved  with  reference 
to  an  expression,  by  regarding  it  as  a  single  letter. 

1.   Solve  the  equation  (x  -f  5)^  -  3  (a;  +  5)^  =  40. 
Completing  the  square, 

Extracting  square  roots,  (a;  -f  5)^  —  |  =  ±  i^- 
,Then,  •(a;  +  6)t  =  |±^  =  8or-5. 


294  ADVANCED   COURSE   IN   ALGEBRA 

We  reject   the   value  (aj  +  5)^  =  — 5;  the  only  solution  is 

{x  +  5)^  =  8. 

Extracting  cube  root,   (x  +  5)*  =  2. 

Raising  to  fourth  power,  a;  +  5  =  16,  and  x  =  11. 

Certain  equations  of  the  fourth  degree  may  be  solved  by  the 
rules  for  quadratics. 

2.    Solve  the  equation  a;*  +  12  »« +  34  a^  -  12  a;  -  35  =  0. 

The  equation  may  be  written 

(a;*  +  12a^  +  36.'^)-2a^- 12x  =  35. 

Or,  (x'^Q>xf-2{x^  +  (^x)  =  Z5. 

Completing  the  square, 

(a^  +  6  0^)2-2(032  +  6  a?) +1  =  36. 

Extracting  square  roots,   (a^  +  6  .t)  —  1  =  ±  6. 

Then,  aj2  + 6  a;  =  7  or  -5. 

Completing  the  square,        x^  +  6  a;  +  9  =  16  or  4. 

Extracting  square  roots,  a;  +  3=±4or  ±2. 

Then,  .a;  =  -3±4  or  -3±2  =  1,  -7,  -1,  or  -5. 

In  solving  equations  like  the  above,  the  first  step  is  to  complete  the 
square  with  reference  to  the  x^  and  x^  terms  ;  by  §  447,  the  third  term  of 
the  square  is  the  square  of  the  quotient  obtained  by  dividing  the  x^  term 
by  twice  the  square  root  of  the  a;*  term. 


3.    Solve  the  equation  x^  —  Qx-\-  B^x"  -  6  a;  +  20  =  46. 

Adding  20  to  both  members, 

(aj2 _ 6  a;  +  20)  +  5 Va;2-6a;  +  20  =  m. 

Completing  the  square, 

(a;2_6a;  +  20)+5Va;2-6a;  +  20  +  ?^  =  66  +  ?^  =  ^. 
^  ^  4  4        4 

Extracting  square  roots,  Vaj^  —  6a3  +  20  +  -  =  ±— -• 


Then,  Va^-6a;  +  20=:6  or  -11. 


EQUATIONS   SOLVED   LIKE   QUADRATICS  295 


The  only  solution  is  Vx^  —  Qx-{-20  =  6. 

Squaring,  a;^- 6  a; +  20  =  36. 

Completing  the  square,  aj^  —  6  a;  -f  9  =  25. 

Extracting  square  roots,  a;  —  3  =  ±  5,  and  a;  =  8  or  —  2. 

In  solving  equations  of  the  above  form,  add  such  an  expression  to  both 
members  that  the  expression  vsrithout  the  radical  sign  in  the  first  member 
may  be  the  same  as  that  within,  or  some  multiple  of  it. 


4.    Solve  the  equation  2x^-\-5x  —  2  xVx^ -f- 5  a;  —  3  =  12. 
The  equation  may  be  written 

a:2  +  5a;-2a;Va^  +  5a;-3  +  a^  =  12.. 

Subtracting  3  from  both  members. 


(a;2  +  5a;-3)-2a;Va^  +  5a;-3  +  a;2  =  9. 
Extracting  square  roots,  VxP  +  5a;  —  3  —  a;  =  ±3. 


Or,  ^  Va^  +  5a;-3  =  a;±3. 

Squaring,  a^  +  5a;  —  3  =  a52±6a;  +  9. 

Then,  -x  or  lla;  =  12,  and  a;  =  -12  or  i?. 
Neither  value  satisfies  the  given  equation. 

5.    Solve  the  equation  ^:=:^  +  ^^::^  =  §. 

X^  —  X        X    —  O        ^ 

a^  — 3 
Kepresenting by  y,  the  equation  becomes 

ai/  —  X 

y  +  -  =  t  or  2/  +  2  =  52/. 

Solving  this,  ?/  =  ^  or  2 ;  that  is,  ^^  =  ^  or  2. 

2  scr  —  x      2 

Taking  first  value,    2  a;^  —  6  =  ar^  —  a;. 
Or,  ar^  +  a;  =  6. 

Solving,  a;  =  2  or  —3. 

Taking  second  value,  a;^  —  3  =  2ar^  —  2a;. 
Or,  -a2^2aj  =  3. 

Solving,  a;  =  1  ±  V— 2. 


296  ADVANCED  COURSE  IN  ALGEBRA 

EXERCISE  60 

Solve  the  following : 

1.  iK*  -  29 a;2  =  -  100.  9.    6x-2  =  nVx. 

2.  0.-6  +  19  ^-s  =  216.  ^^    ^_V-  ^  244  x'^  =  -  243. 

3.  xUl0xJ  +  9  =  0.  11    2x-B-35x-*  +  48  =  0. 

4.  x^-33x^  =  -32.  13^   27x6  +  46x3  =  16. 
6.   64  +  63x-t-x-  =  0.  13    32  ^f  _  33  ^  _  ,-i. 

6.  8x-2  +  14x-i  =  5. 

_6  3  14.    16x8-33x4-243  =  0. 

7.  5x^+7x^=-2. 

2      18  15.    161x5 +  5  =-32x10. 

8.  4x^-  — =  -21. 


X* 

16.    81x~^-308- 

-  64  x^  = 

=  0. 

17. 

(2  x2  +  3  x)2  -  4(2  x2  +  3  x)  =  45. 

18. 

x*  +  12  x3  +  14  x2  _  132  X  -  135  =  0. 

19. 

5x  +  12-5V5x+  12  =  ~4. 

20. 

x2-3        2x    _     17 
2  X        x2  -  3          4 

21. 

3  x2  +  X  +  5V8  :k2  ^  a;  _}_  6  =  30. 

J.   8  x2  -  1  +  6  X  V8  x2  -  1  =  -  8  x2. 

23.  X*  -  2  ax8  -  17  a2x2  +  18  a^x  +  72  a*  =  0. 

24.  lsfx--Y  +  2llx--\=-b. 

25     a;^  +  2      2  X  -  5  _  35 
*2x-5      x2  +  2       6* 

26.   x2  -  6  Vx2  -4x  +  11  =  4 X  -  19. 


27.  x2  +  3  X  +  4  -  Vx2  +  3  X  +  4  =  2. 

28.  (4  x2  +  2  X  -  7)2  +  4  x2  +  2  X  -  189  =  0. 

29.  \/x2  -  3  X  -  3  =  x2  -  3  X  -  23. 

30.  (2  x2  -  3  X  -  1)3  -  6(2  x2  -  3  X  -  1)^  =  16. 
81.  3v^x2  -  12  X  -  7  v'x2  -  12  X  =  -  2. 

32.  x*  -  18  x3  +  109  x2  -  252  X  +  180  =  0. 

33.  V3  x2  -  2  X  +  16  +  2  v/3  x2  -  2  X  +  16  -  16  =  0. 

34.  7(x3  -  28)"^  +  8(x3  -  28)"^  =  -  1. 

35.  (3x  +  15)*  +  9(3x  +  15)*  =  22. 

36.  5  x2  +  6  X  -  166  =  4  x  Vx^  +  5  x-  8. 


EQUATIONS  SOLVED  LIKE   QUADRATICS         297 


37.  10(1  -  x2) -  12  X  -  V5ic2  +  6x-2  =  0. 

38.  x*  +  28  a;3  +  190  x2  -  84  x  ^  135  =  0. 

39.  9(x  +  a)^  -  22  b^(x  +  a)^  +  8  6*  =  0. 


40.  x2  +  l+Vx2-8x  +  37  =  8(x  +  12). 

41.  25(x  +  l)-i  -  15(x  +  1)"^  =  -  2. 

42.  (3  X  -  2  a)2  -  4  x(3  X  -  2  a)  =  4(x  +  4  a)2  +  8  x(x  +  4  a). 

5c2_5a;  +  i      a;2_2x  +  2         8 


43 


2x  +  2     x2-5x  +  l         3 


44.   9  x(7  -  x)  +  9\/x2  _  7  X  -  5  =  -  43. 

45    3  x2  +  2  X  -  5     4  x2  -  7  X*-  1  ^  5 
•4x2_7x-13x2  +  2x-5     2* 

46.   9  X*  -  30  x3  -  185  x2  +  350  x  +  1176  =  0. 


47       /x2  +  3x+i0        /x2-5x  + 
\x2-5x  +  2  "^  \x2  +  3x  + 


48.      "^    ■  "        '     -^  ^ 


10      5* 


V^^-aI 


x2  +  3     2 


298  ADVANCED  COURSE  IN   ALGEBRA 


XXI.    SIMULTANEOUS    QUADRATIC 
EQUATIONS 

On  the  double  signs  ±  and  :p. 

If  two  or  more  double  signs  are  used  in  an  equation,  it  will  be  under- 
stood that  the  equation  can  be  read  in  two  ways  ;  first,  reading  all  the 
upper  signs  ;  second,  reading  all  the  lower  signs. 
Thus,  the  equation  a  ±b  =  ±c  can  be  read  either 

a  -\-b  =  c,  OT  a  —  b  =  —  c. 
And  the  equation  a±b  =  =p  c  can  be  read  either 
a  +  6=—  c,  or  a  —  5  =  c. 

The  same  notation  will  be  used  in  the  case  of  a  system  of  equations, 
each  involving  double  signs. 

Thus,  the  equations  x  =  ±2,  y  =  ±S,  can  be  read  either 

X  =  +  2,  2/  =  +  3,  or  a;  =  -  2,  y=-3. 
And  the  equations  x=  ±2,  y  =  =p3,  can  be  read  either 

aj  =  +  2,  y=-S,  or  X  =  -  2,  y=+S. 


The  principles  demonstrated  in  §§  233  to  236,  inclusive,  hold  for  simul- 
taneous equations  of  any  degree. 

470.   The  following  principle  is  of  frequent  use  in  solving 
simultaneous  equations  of  higher  degree  than  the  first : 
The  system  of  equations 

AxB  =  0,  (1) 

CxD  =  0,  (2) 

where  A,  B,  C,  and  D  are  rational  and  integral  expressions 
which  involve  the  unknown  numbers,  is  equivalent  to  the 
systems 

lo=o,        [d=o,        \c=o,        [d=o. 

For  any  solution  of  (1)  and  (2)  makes  Ax  B  and  G  X  D 
identically  equal  to  0. 


SIMULTANEOUS   QUADRATIC   EQUATIONS        299 

It  then  makes  at  least  one  factor  oiAxB  and  C  x  D  identi- 
cally equal  to  0,  and  hence  satisfies  some  one  of  the  systems  (3). 

Again,  any  solution  of  any  one  of  the  systems  (3)  makes 
either  A  or  B,  and  also  either  C  or  D,  identically  equal  to  0 ; 
and  hence  satisfies  (1)  and  (2). 

Then,  the  system  (1)  and  (2)  is  equivalent  to  (3). 

The  principle  holds  for  any  number  of  equations,  with  any  number  of 
factors. 

471.  Two  equations  of  the  second  degree  (§  113)  with  two 
unknown  numbers  will  generally  produce,  by  elimination,  an 
equation  of  the  fourth  degree  with  one  unknown  number. 

Consider,  for  example,  the  equations 

V-f2/=a.       .  (1) 

.x-^f=b.  (2) 

From  (1),  y  =  a  —  oi^;  substituting  in  (2), 
a;  +  a^  —  2  ax^  -f  a;*  =  6 ; 
an  equation  of  the  fourth  degree  in  x. 

The  methods  already  given  are,  therefore,  not  sufficient  for 
the  solution  of  every  system  of  simultaneous  quadratic  equa- 
tions, with  two  unknown  numbers. 

In  certain  cases,  however,  the  solution  may  be  effected. 

472.  Case  I.    WJien  each  equation  is  in  the  form 

ax^  +  by^  =  c. 

In  this  case,  either  x^  or  y^  can  be  eliminated  by  addition  or 
subtraction. 

3^2+   42/2  =  76.  (1) 


1.    Solve  the  equations  ,       „     ^      „ 

^  '32/2-11x2  =  4.                              (2) 

Multiply  (1)  by  3,  9x'  +  12y^  =  228. 

Multiply  (2)  by  4,  l2/-44a^=   16. 

Subtracting,  53  a^  =  212. 

Or,  a^=     4.                           (3) 

Whence,  x  =±2, 


300  ADVANCED  COURSE  IN   ALGEBRA 

Substituting  a;  =  ±  2  in  (1),  12  +  4  /  =  76. 

Then,  /  =  16.  (4) 

Whence,  2/  =  ±  4. 

The  solution  is  a;  =  2, 2/  =  ±  4  j  or,  a;  =  —  2, 2/  =  ±  4.  (5) 

It  follows,  precisely  as  in  Ex.  1,  §  238,  that  the  given  system  is  equiva- 
lent to  the  system  (3)  and  (4). 
Now  (3)  and  (4)  may  be'written 

(x  +  2) (cc  -  2)  =  0,  and  (y  +  4)(?/  -  4)  =  0. 
And  by  §  470,  these  are  equivalent  to 

which  are  the  same  as  (5). 

The  method  of  elimination  by  addition  or  subtraction  may  be 
used  in  other  examples. 

Sx'-4.y=    ^7.  (1) 

62/=   33.  (2) 

Multiply  (1)  by  3,  9  x^  -  12  ?/  =  141. 

Multiply  (2)  by  2,      '   Ux'^12y=   66. 
Adding,  23  a:^  ^  207. 

Then,  o^  ^  9^  and  a;  =  ±  3. 

Substituting  a;=  ±3  in  (1),  27-42/  =  47. 

Then,  -  4  2/  =  20,  and  2/  =  -  5. 

It  is  possible  ^o  eliminate  one  unknown  number,  in  examples  (1)  and 
(2),  by  substitution  (§  239),  or  by  comparison  (§  240). 

EXERCISE  61 

Solve  the  following : 

3  ic2  +  2  y2  =  66.  ^     r  4  a;2  +  9  2/2  _  13. 


r3a:2_ 

2.   Solve  the  equations  \ 

^  I7aj2  + 


■  9  x2  +  5  2/2  =  189.  I  8  x2  -  27  2/2  =  6. 

g     ,  3a;- 52/2  =  _  116.  ^     (^xy+     y'^=-75. 

7x  + 42/2  =  121.  '1    a-y  -  3 2/2  =  -  95. 

3  a;2  -  2  X2/ =;  24.  j  11  a;2  -    6  2/2  =  84. 

4  a;2  -  5  a;2/  =  46.  '*    L   7  x2  +  15  y^  =  204. 


2/2  +  4  a;?/  -  3  ?/  =  42. 
2 2/2  _    xy  +  6y=- 10. 


SIMULTANEOUS  QUADRATIC  EQUATIONS         301 

(2x^-xy-Sy'^  =  0.  (2x^  +  3y^  =  61  -  x. 

'   \    ^2  +  xy  +  3  2/2  =  27.  ■  '1    ic2  -  2  2/2  =  17. 

^     r4x2-2/2=(3a+&)(a+36).        ^^ 
l4  2/2-x2=(3a-6)(a-3&). 

(Sx  +  2y  Sx-2y^41 
11    J  3  X  -  2  2/     3  iK  +  2  2/     20' 

[  8  2/2  +  3  x2  =  29. 

^g     r  (3a: +  42/)2- (6a:-    2/) (6 x  +  5 ?/)  =  57. 

1  (5  a:  -  2  2/)2  -  4(x  -  4  2/)  (2  a:  +  3  2/)  =  225. 

473.  Case  II.  When  one  equation  is  of  the  second  degree, 
and  the  other  of  the  first. 

Equations  of  this  kind  may  be  solved  by  finding  the  value  of 
one  of  the  unknown  numbers  in  terms  of  the  other  from  the 
first  degree  equation,  and  substituting  this  value  in  the  other 
equation. 

Ex.     Solve  the  equations      [2 a^' -  a^2/  =  6 y.  (1) 

1     X  +2y=7.  .     (2) 

From(^,  2/  =  ^-  (3) 

Substituting  in  (1),        2x'-x  f^^)  =  6  C^^^) '  (4) 

(5) 


Clearing  of  fractions, 

4a 

^- 

-7x  +  x'  = 

=  42- 

■6  a;. 

Or, 

^x'-x^ 

=  42. 

Solving, 

x  = 

=  3  or 

_I4 

Substituting  in  (3),  y  = 

7- 
=     2 

3 

or       ^      = 

=  2  or 

49 
10* 

The  solution  is  x  =  3,  y  =  2  -,  or,  x  =  — —,  y  =  —-.  (6) 

5  10 

By  §  236,  the  given  system  is  equivalent  to  (3)  and  (4)  ;  or,  since  (4) 
is  equivalent  to  (5),  to  the  system  (3)  and  (5). 

Now,  (5)  can  be  written  (x  —  3)(5x  +  14)  =  0. 

Then,  by  §  470,  the  system  (3)  and  (5)  is  equivalent  to  the  systems 
(3)  and  X  -  3  =  0,  and  (3)  and  5  x  +  14  =  0  ;  that  is,  to  (6). 


302 


ADVANCED  COURSE  IN   ALGEBRA 


EXERCISE  62 

Solve  the  following : 

\x  -2y  =9. 

9. 

'-r-f 

1            xy  =  2  a2  +  3  a  -  2. 
\Sx  +  4:y  =  na-\-2. 

^+!=¥-  ■ 

"^      I* 

(4x  +  y-Sxy  =  -6. 

(2xy  +  x  =  ~S6. 
ixy  —  Sy  =  —  6. 

10. 

\x-6y  +  2xy  =  10. 

(x^-2y^  +  Sx=-S. 

11. 

X       ^  —  y  _    40 

X  -  w         X             21 

[.x^-2y^-4y  =  -2. 

2y+3x  =  -l. 

(x^-xy  +  2y'^  =  8. 
'   \             Sx  +  y  =  10. 

12. 

'   X        ?/   _29 
.  2y      Sx      24' 

(Sx^-xy-y^  =  -S. 

4?/-x=-2. 

L          2x-Sy  =  6. 

ft;2  +  4  x?/  =  13. 
.  2  x?/  +  9  y^  =  87. 

\2x     Sy_^ 
3^2 

13. 

'*   1-  +  ^-^- 

r  ^-1    ^-1  =  0. 

1         3             2 

^  2  X     Sy 

14. 

. 

a;2  _  y       y^  -X       Q 

I     32             15           ' 

.^f-^-l 

15. 

'  x'^y^  -  24  x?/  +  95  =  0 
3  X  -  2  ?/  =  - 

I             X  +  ?/  = 

(2 

a  — 

&)(a  +  7  6). 

:3( 

a  — 

&) 

. 

13. 


474.  Case  III.  When  the  given  equations  are  symmetrical 
with  respect  to  x  and  y ;  that  is,  when  x  and  y  can  he  inter- 
changed without  changing  the  equation. 

Equations  of  this  kind  may  be  solved  by  combining  them  in 
such  a  way  as  to  obtain  the  values  oi  x-\-y  and  x  —  y. 

[x  +  y  =  2.  (1) 

1       xy  =  -W.      .  (2) 

Squaring  (1),  o^  +  2xy  +  f-=       4.  (3) 

Multiplying  (2)  by  4,  4  xy         =  -  60.  (4) 

Subtracting,  x^  —  2xy-\-y^=     64.  (5) 

Extracting  square  roots,  x  —  y  =  ±S.  (6) 


1.   Solve  the  equations 


SIMULTANEOUS   QUADRATIC   EQUATIONS         308 

Adding  (1)  and  (6),  2  a;  =  2  ±  8  =  10  or  -  6. 

Whence,  ic  =  5  or  —  3. 

Subtracting  (6)  from  (1),  2  ?/  =  2  t  8  =  -  6  or  10. 

Whence,  2/  =  —  3  or  5. 

The  solution  is  a;  =  5,  2/  =  —  3 ;  or,  a;  =  —  3,  ?/  =  5.              (7) 

In  subtracting  db  8  from  2,  we  have  2  ^  8,  in  accordance  with  the  nota- 
tion explained  on  page  298. 

In  operating  with  double  signs,  ±  is  changed  to  T ,  and  =F  to  ± ,  when- 
ever +  should  be  changed  to  — . 

Equation  (4)  is  equivalent  to  (2)  ;  but  (3)  is  equivalent  to  x-\-  y  =  2 
and  X  -\-  y  =  —  2. 

If,  then,  we  use  only  the  value  +2  tor  x+  y,  the  given  system  is 
equivalent  to  (3)  and  (4). 

By  §  234,  the  system  (3)  and  (4)  is  equivalent  to  (3)  and  (5). 

Then,  the  given  system  is  equivalent  to  the  positive  value  oi  x  -{■  y  in 
(3),  and  (5)  ;  that  is,  to  (1)  and  (5). 

Now  (5)  may  be  written  {x  —  y  —  8)  (x  —  y  +  S)  =  0. 

Then,  by  §  470,  the  system  (1)  and  (0)  is  equivalent  to  (1)  and 
X  —  y  —  S=0,  and  (1)  and  x  —  y  +  S  =0;  that  is,  to  (7). 


The  above  equations  may  also  be  solved  by  the  method  of  Case  II ;  but 
the  symmetrical  method  is  shorter  and  neater. 


2.    Solve  the  equations 


a;2  4- 1/2  =  50. 
xy  =  -7. 

Multiply  (2)  by  2,  2xy  =  - 14. 

Add  (1)  and  (3)  x" -\- 2  xy -\- y' =  36. 

Whence,  x-\-y=±6. 

Subtract  (3)  from  (1),  x^-2xy-{-y^ 

Whence, 

Add  (5)  and  (7), 

Whence, 

Subtract  (7)  from  (5), 

Whence, 

The  solution  is  x=  ±7,  ?/  =  =F  1 ;  or,  a;  =  ±  1,  i/  =  T  7. 


=  ±6 
=  64.     J 


X 

-y  = 

-.±S.^ 

2x  = 

=  6  ±  8,  or 

-6±8. 

x  = 

=  7,-1,1, 

or  -7. 

2y  = 

^  6  T  8,  or 

-6q=8. 

?/  = 

=  -1,7,- 

7,  or  1. 

304 


ADVANCED  COURSE  IN  ALGEBRA 


The  given  system  is  equivalent  to  the  system  (4)  and  (6). 
Now  (4)  and  (6)  can  be  written 

(X  +  y  -{-  6){x  +  y  -  6)  =  0,  a^nd  (x  -  y  +  S)(x  -  y  -S)=0. 

Then,  the  system  (4)  and  (6)  is  equivalent  to  (5)  and  (7),  with  every 
possible  combination  of  signs. 


We  may  solve  by  the  method  of  Case  III  other  systems  in  which  the 
equations  are  symmetrical,  except  in  the  signs  of  terms  ;  as,  for  example, 

the  system 

fx-y  =  a. 

\       xy  =  b. 

We  may  also  solve  certain  non-symmetrical  systems ;  as,  for  example, 

the  system 

r  a2x2  +  62^2  ^  c. 

[      ax+  by  =  d. 


EXERCISE  63 

Solve  the  following : 

^    |x2  +  2/2  =  29. 

'x^-\-xy  +  y^=:^ 

I    x  +  y  =  -3. 

9. 

4 

lx-y  =  n. 

^'    I       xy  =  -2S. 

x^-xy-^-y-^^^j- 

^     ra;2  +  2/2  =  i3o. 
*  1    x-y  =  -8. 

10. 

X^       2/2 

1-1=12. 

(x-^y  =  2a-l. 
'  \      xy  =  a'^-  a-2. 

I    X     y 

f  1       1  _  289 
x^     y^      36 

1  _io. 

..  {'-'•='^- 

11.  ^ 

I         xy  =  ab. 

xy      3* 

'I            x-y  =  S. 

12.  ^ 

'a;2  +  9?/2  =  50. 
,    x-Sy  =  0. 

r?  +  |/  =  -12. 

13. 

xy  =  -16. 
.  2  X  +  2/  =  14. 

1.  .  y     X         3 

[x-y  =  l. 

r       1 

(x'^-xy  +  y^  =  a^  + 
'   [           x-hy  =  2a. 

3  62. 

14.  ^ 

-  =  24. 

xy 

x  +  y  =  llxy. 

SIMULTANEOUS  QUADRATIC  EQUATIONS        806 


15. 


16. 


1  +  i  +  l 

x2     xy     y^ 


49. 


1  +  1  =  8. 

X     y 


r36x2  + 
I   6x  + 

-I 


x^  —  xy  -{■  y2 
3:2^/2 


iC?/ 


_7_ 
324* 
J_ 
18* 


21. 


l  +  i.  +  l  = 

r2     icy     y2 


_i_  +  l  =  ^ 


8y  =11. 
25  x2  +  16  y2  =  544. 
5x  -    4  2/  =32. 
r  2  x2  -  3  icy  +  2  y2  =  92. 
'    15x2 +  4  a;?/ +  5^2 -161. 

16  x22/2  _  104  xy  =  -  105. 
X  —  y  =  —  2. 
3  g^  -  3  a262  +  54 
a2(a  -  6)2 
4  _  ^262  +  54 


■■I 


x?/     y^        a^^a  —  &)2 

475.  Case  TV.  When  each  equation  is  of  the  second  degree, 
and  homogeneous /  that  is,  when  each  term  involving  the  unknown 
numbers  is  of  the  second  degree  with  respect  to  them. 

Certain  equations  of  this  form  may  be  solved  by  the  method  of  Case  I 
or  Case  III.  (See  Exs.  1,  §  472,  and  2,  §  474.)  The  method  of  Case  IV 
should  be  used  only  when  the  example  cannot  be  solved  by  Cases  I  or  III. 

Ex.     Solve  the  equations      \    „  „~   ' 

Putting  in  the  given  equations  y  =  vx, 

5 


we  have  a^  —  2vx^=   5 ;   or,  aj^  = 

and  0^  +  1)^0^  =  29',   or,  0^  = 

Equating  values  of  a^, 

Or, 

Solving  this  equation. 


29 

5      ^    29 
l-2v     l  +  u^' 

5'u2+58'y  =  24. 


(1) 
(2) 


(3) 

(4) 
(5) 


-y  =  -  or  —  12. 
5 


Substituting  these  values  in  (2),  we  have 


x" 


or 


1- 


1  +  24 


25  or  9;    then,  x  =  ±5  or   ± 

5  V6 


306 


ADVANCED   COURSE   IN   ALGEBRA 


Substituting  the  values  of  v  and  x  in  the  equation  y  =  vx, 


|(,5)or-12(.^) 


±2  or   T 


1?. 


The  solution  is  x=  ±5,  y 


1  12 

±  2 ;  or,  a;  =  ±  — z,  2/  =  T  -— • 


The  given  system,  and  (1),  are  three  equations  with  three  unknown 
numbers  ;  by  §  236,  they  are  equivalent  to  the  system  (1),  (2),  and  (3). 

Then,  precisely  as  in  Ex.,  §  240,  the  system  (1),  (2),  and  (3)  is 
equivalent  to  (1),  (2),  and  (4),  or  to  (1),  (2),  and  (5). 

We  may  write  (6)  in  the  form  (6v  —  2){v  +  12)  =  0. 

Then,  the  system  (1),  (2),  and  (5)  is  equivalent  to  the  systems  (1), 
(2),  and5v-2  =  0,  and  (1),  (2),and  v+  12  =  0. 

Then,  the  given  system  is  equivalent  to  the  systems 


and  y  z=—12x,  x"^ 


1- 


1  +  24 


In  finding  y  from  the  equation  y  =  vx,  care  must  be  taken  to  multiply 
each  value  of  x  by  the  value  of  v  which  was  used  to  obtain  it. 


Solve  the  following : 
25. 
xy  =  4. 
+  3  xi/  =  -  5. 


EXERCISE  64 


1^2+   2/2 

Ix"^  —  XV 


'^1 


2xy  -y^  =  -  24. 
x^  +  xy  +  y^  =  19. 
2x'^  +  xy  =  -2. 


( ix^  —  xy  —  y^ 
L         Sxy  +  y^ 


16. 


28. 


^( 


A-l  =  9. 

a;2      2/2 

±.-^  =  -90, 
xy     2/2 

x2  -f-  ojj/  -  5  2/2  =  25, 
a;2  +  4  2/2  ^  40. 


7. 


10. 


11. 


IB.  \'^- 


a;2  -  2  a:2/  -  4  y2  =  _  41. 

ic2  -  5  ici/  +  8  2/2  =  58. 

r  2  ic2  +  7  iC2/  +  4  y2  =  2. 
1  3  x2  +  8  X2/  -  4  y2  =  _  72. 

r  11  x2  -    xy-    2/2  =  45. 
I   7  a;2  +  3  £C2/ -  2  2/2  =  20. 

5  ic2  +    xy-3y^  =  27. 

4  «2  _  4  x2/  +  3  2/2  =  72. 

x2     xy     y^ 

i-i  =  -12. 

xy     y^ 

4  ic2  -  2  iC2/  -    2/2  =  -  16. 
4  a;2  +  7  a;2/  +  2  2/2  =  104. 
xy  -  40  x2  =  30  x^y^. 
Sxy  -  72  ic2  =  38  x^y^. 


12. 


SIMULTANEOUS   QUADRATIC   EQUATIONS         307 

476.   Solution  of  Simultaneous  Equations  of  Higher  Degree  by 
Factoring. 

1.    Solve  the  equations  \^       ^n   n-    j       ;    . 

[(x  +  y){x-Sy  +  2)=0. 

By  §  470,  the  given  system  is  equivalent  to  the  systems 


x-y  =  0, 

x  +  y  =  0, 


x-y  =  0, 
[x-Sy-{-2  =  0, 


(x  +  2y-l  =  0,     ^^^      cx-^2y-l  =  0, 

I  xi-y  =  0,    ^"^       \x-3y  +  2  =  0. 

The  solutions  of  these  are 

x  =  0,y  =  0',  x  =  l,  y  =  l;  x  =  -l,  y  =  l;  x  =  --,  y  = 

o 


2.   Solve  the  equations  | 


x'-^xy-2y^  =  0.  (1) 

2x4-32/ +  6  =  0. 

We  may  write  (1)  in  the  form  {x  -]-  2  y)(x  —  y)  =  0. 
Then  the  given  system  is  equivalent  to  the  systems 

r  x  +  2y  =  0,    ^^^    I  x-y  =  0, 

\2x-{-Sy-{-6  =  0,   ^^      [2x  +  3y-^6  =  0. 

/»  f* 

Solving  these,  x  =  —  12,  y  =  6;   or,  x  =  —  -,  y  =  —  -' 

5  5 

The  example  can  be  solved  by  the  method  of  §  473  ;  but  the  above 
method  is  shorter. 

Sx^+f-=7x.  (1) 

xy-^f=2x.  (2) 

Multiplying  (1)  by  2,     6  a^+2  /=14  x. 
Multiplying  (2)  by  7,    7  xy-\-7y^= 14  x. 

Subtracting,  6x^-7  xy-5y^=0,  or  (2x-\-y)(Sx-5y)=0. 

Then,  the  given  system  is  equivalent  to  the  systems 

(Sx'-^y'  =  7x,      ^^^      (Sx'-h    f  =  7x, 

120?  +2/  =0,         ^^        l3a;-52/=0. 

25  5 

Solving  these,  a;  =  0,  y  =  0 ;  a;=l,  2/  =  —  2 ;  or,  ic  =  — ,  2/  =  j* 


3.    Solve  the  equations  \ 


308  ADVANCED  COURSE  IN  ALGEBRA 


4.    Solve  the  equations    j    ^ 


a^=    x  +  y. 
y^=3y-x. 

Subtracting,  x^  —  y^=2x—2y,OT{x—y)(x-\-y—2)=0. 

a^  =  x  +  y, 


Then 


,    I       ^  =  ^  +  ^'      and      j  ^  =^  +  ^^ 

Solving  these,   x  =  0,   y  =  0]    x  =  2,   y  =  2)    or,   x=± V2, 

y  =  2T-^2. 

EXERCISE  65 

Solve  the  following : 

(2x-3?/)(a;-4?/  +  l)  =  0.  ^  j  (x  +  2  ?/)(4ic  -  3y)  =  0. 

5  a;  +  4  ?/  =  23.  '  L      3  x2  -  5  a:?/  -  4  ?/2  =  18. 

^     r(2x+     2/)(3a:-4?/  +  5)  =  0.  ^  j  ^^^  _  4  ^2  ^  3a;. 

*  1(   x-3?/)(2x  +  5y-8)=0.  '  1  a:?/ +  8  2/2  =  X. 

:,2_3^y_4y2^0.  10  j3^'  +  ^2/-22/2  =  -4y. 

o„        r..  .«  1 6x2 -2X2/ +  1/2  =  3?/. 

r  3  x2  +  3  ?/2  =  10  xy. 

1  +  1=.! 
X     v     3 


( 


«,.i..        "■       hi-' 


J  2  x2  -  xy  -  15  2/2  =  0. 

1  6x  + 

^     r  12x2  -  11  x?/  +  2  2/2  =  0.  I  (a;  _  y)(3.2  _  sc?/  -  6  2/2)=  0. 

•   1  7x-32/  =  -5.  ^2-  t  2x-5v  +  l  =  0. 


r  2  x3  +  5  xs?/  -  8  X2/2  -  20  ^f  =  0. 
3x  +  42/  =  7. 


x2  =  mx  —  wy. 
xy  -\-y'-  =  44.  ly"^  =  nx  —  my. 


J  2  x2  -  xy  =  0.  r 

^-   1x2  +  22/2  =  9.  ^^-  1 

7     r  x2-2xy  =  0.  ^^ 

I  2  x2  +  xy  +  2/2  =  44. 

477.   Solution  by  Division. 

n      'A     .1.  ,'         {AxB  =  axb,  (1) 

Consider  the  equations  j  ^^     , 

I  B  =  b',  {z) 

where  A  and  B  are  rational  and  integral  expressions  which 
involve  the  unknown  numbers,  and  a  and  b  any  numbers. 
B7  §  236,  the  given  system  is  equivalent  to 

Axb  =  a  xb,  ( A  =  a,  (3) 

_     ,  or,  to  j  ^     , 


SIMULTANEOUS   QUADRATIC   EQUATIONS         309 

Here  the  first  member  of  (3)  is  obtained  by  dividing  the  first 
member  of  (1)  by  the  first  member  of  (2),  and  the  second  mem- 
ber by  dividing  the  second  member  of  (1)  by  the  second  member 
of  (2). 

(a^-y^  =  9.  (1) 

Ex.   Solve  the  equations  \ 

^  [   x-y  =  S.  (2) 

Divide  (1)  by  (2),        a^  +  xy-{-f  =  3.  (3) 

Squaring  (2) ,  a?~2xy  +  /  =  9.  (4) 

Subtract  (4)  from  (3),  3  x?/  =  -  6,  or  0*2/=  -  2.         (5) 

Add  (3)  and  (5),      a^  +  2  a;?/  +  /  =  1- 

/  Extracting  square  roots,        x-{-y=±l.  (6) 

Add  (2)  and  (6),  2a;  =  3±l=4or2. 

Then,  aj  =  2orl. 

Subtract  (2)  from  (6),  22/=-3±l=-2or-4. 

Then,  y=-lov-2. 

The  solution  is  a?  =  2,  2/  =  —  1 ;  or,  a;  =  1,  2/  =  —  2. 

The  equations  (2)  and  (3),  though  not  symmetrical,  are  solved  as  in  §  474. 

478.   Consider  the  equations  \ 

where  A,  B,  C,  and  D  are  rational  and  integral  expressions 
which  involve  the  unknown  numbers. 

By  §  236,  the  given  system  is  equivalent  to 

(AxB=CxB,  ^     (B(A-C)  =  0, 

i  or,  to 

1         B==D;  '        1  B  =  D. 

By  §  470,  the  latter  is  equivalent  to  the  systems 

(B  =  0,  ,    (A-C  =  0, 


{d=d, 


and 


B  =  D. 


Then,  the  given  system  is  equivalent  to  the  systems 
li)  =  0,  [b=d. 


310 

Ex.    Solve  the  equations 


ADVANCED   COURSE  IN   ALGEBRA 


(1) 

x-y  =  3x-\-2.  (2) 

Dividing  (1)  by  (2),  the  given  system  is  equivalent  to  the 
systems 

x-{-y  =  9x^  —  6x-{-4:j 


x-y  =  0, 
3  a;  +  2  =  0, 


and 


( x-{-y  =  9  or 
[  x  —  y  =  S X 


+  2. 


2              2 
The  solution  of  the  first  system  is  x= ,  y= . 

o  o 

To  solve  the  second,  add  (2)  and  (3) ;  then, 

2x=9o^-Sx-\-6,ov9x~-5x-\-6  =  0, 
5  ±  V25-216     5  ±  V-m. 


By  §  450, 

By  (2), 


x  = 


18 


18 


y=^2x-2=-^^^-^^^-2 
^  9 

^-23q:V-191 

9 


(3) 


If  we  try  to  solve  by  substituting  the  value  of  y  from  (2)  in  (1),  we  shall 
have  an  equation  of  the  third  degree  in  x. 


Solve  the  following : 

a;3  -  2/3  _  26. 

x-y  =  2. 

x^  +  y^  =  280. 
x^-xy-^y^  =  28. 
r        x  +  y  =  S5. 
I  v^a;  4-  v^  =  6. 
=  189. 

y 


EXERCISE  66 


ra:3_8?/3r=U 
'I    X  -  2  V  =  9. 


flj3       2/8 

X     y 


l+±  +  l  =  12. 

I  x^     xy      2/2 


1         X 


9. 


10 


+  x^2  _  56. 
+  ?/  =  -l. 
aJ*  +  «22/2+y4  =  481. 
x2  -  xy  +  2/2  =  37. 
r(x2-16)(9  2/2-4)=-2100. 
I      (x  +  4)  (3  2/  -  2)  =  60. 
r  25  x2  -  25  2/2  =  24  x22/2. 
I       5x  +  5y  =  -4x2/. 


SIMULTANEOUS   QUADRATIC   EQUATIONS         311 


11. 


12. 


13. 


r  8  a;3  +  27  y^  =  91. 

U  .x2  _  6  xy  +  9  2/2  =  13. 

jx^-y^  =  ic2?/2  -  1. 

I  X2  —  y2  —  a;?/  —  ] . 


rx2 


i  2/   ■  a;       6* 

1   1  +  Ul. 

I    X     y     6 


14. 


15. 


16. 


2/       x      15 
^    a;-2/  =  2. 

a;2  _  ^2  +  3  a-  _  3  y  ^  12. 
(x-2/)(3x-52/-4)=-9. 
x'^y  +  ?/2x  =  42. 

1+1=1. 

X     y     6 


479.   Special  Methods  for  the  Solution  of  Simultaneous  Equa- 
tions of  Higher  Degree. 

r     Q^ y^  z=z  19. 

1.    Solve  the  equations  j    „  on' 

[  ocry  —  xy'  =  o. 

Multiply  (2)  by  3,  3x'y-3xy'  =  18. 

Subtract  (3)  from  (1),  a^ -Sx'y +3  xy^ -  f  =  1, 

Extracting  cube  roots,  x  —  y  =  l. 

Dividing  (2)  by  (4),  xy  =  6. 

Solving  equations  (4)  and  (5)  by  the  method  of  §474,  we 
find  a;  =  3,  2/  =  2  ;  or,  x  =  —  2,  ?/  =  —  3. 


(1) 

(2) 
(3) 

(4) 
(6) 


«     o.  ,        T  .  (a^  +  y^  =  9xy. 

2.    Solve  the  equations  i       ,  n 

^  [x  -{-y  =6. 


Putting  x  —  u-\-v  and  y  =  u  —  v, 

(u  4-  vy  +  (u  -  vf  =  9(u -\.v)  (u-v);  (1) 

and  (u  +  v)  +  (w  -  v)  =6.  (2) 

Reducing  (1),  2 1^^  +  6  ^^2  ^  9(^^2  _  ^2^^  (3^ 

Reducing  (2),  2  w  =  6,  or  w  =  3. 

Putting  w  =  3  in  (3),  54  + 18  v^  =  9(9  -  v"). 

Whence,  i;^  =  1,  or  v  =  ±  1. 

Therefore,  a;  =  w  +  'V  =  3±l  =  4or2; 

and  2/  =  w  —  'y  =  3Tl  =  2or4. 

The  artifice  of  substituting  u  +  v  and  u  —  v  for  x  and  y  is  advantageous 
in  any  case  where  the  given  equations  are  symmetrical  (§  474)  with 
respect  to  x  and  y.     See  klso  Ex.  4. 


812  ADVANCED   COURSE  IN  ALGEBRA 

aj2  +  2/'  +  2a;  +  22/  =  23. 


3.    Solve  the  equations 

xy  =  6. 

Multiplying  (2)  by  2,  2  xy  =  12. 

Add  (1)  and  (3),  a?-^2xy -\-y'' -\-2x  +  2y  =  Z5. 

Or,  (a;4-2/)2  +  2(aj  +  2/)  =  35. 

Completing  the  square,     {x  +  yY  +  2(x  +  ?/)  + 1  =  36. 
Then,        (a;  +  2/)  + 1  =  ±  6 ;  and  a;  +  ?/  =  5  or  —  7. 
Squaring  (4),  a^  -4-  2  a;?/  +  2/^  =  25  or  49. 

Multiplying  (2)  by  4,  4  a??/         =  24. 

Subtracting,  y? —  2xy-\-.y^—   1  or  25. 

Whence,  a;  —  2/=±lor  ±5. 

Adding  (4)  and  (5),  2  ic  =  5  ±  1,  or  —  7  ±  5. 

Whence,  x  =  S,2,  —  1,  or  —  6. 

Subtracting  (5)  from  (4),  2  ?/  =  5  ^F  1,  or  —  7  q=  5. 

Whence,  2/  =  2,  3,  -  6,  or  - 1. 

fx'^  -\-y^  =:  97. 
aj+2/=-l. 
Putting  x  =  u-{-v  and  y  =  u  —  v, 

(u  +  v)^  +  (w  -  'y)^  =  97, 
and  (u-\-v)  +(u  —  v)  = —1. 

Reducing  (1),     2  ?^^  + 12  u^v'-  +  2v^  =  97. 
Reducing  (2),  2w=— 1,  orw=—  -. 

Substituting  in  (3),    1  -f  3  v^  _^  2  ^*  =  97. 
8 


Solving  this,  v»  =  ?^  or  -51;  and  -1;=  ±  ?  or  ±  ^^nB. 
£>        ^  4  4  '  2  2 

Therefore,  a;  =  w +  i;  =  - -  ±  ^,  or  _!  ^^  V-31 

'    ,    '     ^  2     2'  2  2 


=  2,-3,or-^^y-^^. 


SIMULTANEOUS   QUADRATIC  EQUATIONS         313 


A    ^  l-r^  l^V-31 

And,  y  =  u-v  =  -^T-^ov  --T—^ — 


In  solving  fractional  simultaneous  equations  of  higher  degree,  we  must 
reject  any  solution  which  satisfies  the  equation  obtained  by  equating  to 
zero  the  L.  C.  M.  of  the  given  denominators  (§  222). 

Also,  in  solving  simultaneous  equations  of  higher  degree  having  un- 
known numbers  under  radical  signs,  we  retain  only  those  solutions  which 
satisfy  the  given  equations,  when  the  principal  values  of  the  roots  are 
taken  (§  397). 

EXERCISE  67 

Solve  the  follow  ing  : 

{x'^  -Vy^-  257.  r    a;2y  -  a:  =  -  6. 

y    x  —  y  —  b..  I  x^y^  _  aj3  _  _  72. 

x2  +  2/2  +  a;  _  y  =  32.  r  X  -  ?/  +  v^  =  11. 

^y  =  6.  •^'   1  ^Wy  -  Vx^  =  30. 

J     x3  +  ?/3  =  2a3  +  24a.  <x^^y'^  =  xy-\-\. 

Ix2y  +  x?/2  =  2a(a2_4).  ^^'    [  a;*  +  2/4  =  x2y2  +  1. 

x2  +  9?/2  +  4x  =  9.  r                          a;  +  2/  =  -2. 

xy  +  2?/  =  -2.  ^^'  t  a:5  +  a;8?,2  +  a;2^ ^.  2^  =  _  1120. 

a;  +  y  +  a;y  =  11.  r         a;y  +  x  +  ?/  =  169. 

14. 


'I 


(X  +  yy  +  x?2/2  =  61.  "•    t  V^  _  Vi^  =  7. 

I     x2?/-x?/2  =  -20.  ^^-  t      a;yVx2  +  ?/2  =  _60. 

r9x2-13x2/-3x  =  -123.  f  x2  (1  +  y)  +  yHl  +  a^)  =  109. 

*    I      xy  + 4  2/2  +  2?/ =  125.  ^^-  I  xy  =  12. 

I     V2  x2  -  9  =  3  2/  +  6.  rx5-2/^  =  211. 

[  Vx4  _  17  2/2  =  aj2  _  5.  '''•  I    a._^  =  i. 

x2  +  2/  =  -3.  j(x3  +  2/^)(a;  +  2/)  =  112. 

33.         *    I  x2  +  X2/ +  2/^^  =  13. 

.  2  x32/ -  6  x22/2  +  2  xy3  ^  ^4  =  0. 
19.  i 

X  4-  2/  =  «2/. 


j«*  + 


V2  x2  +  6  2/  +  10  =  19  -  x2  -  3  2/, 
x2+^2_4j,  =  i^ 


314  ADVANCED   COURSE   IN   ALGEBRA 

SIMULTANEOUS   QUADRATIC  EQUATIONS  WITH  MORE 
THAN  ONE  UNKNOWN  NUMBER 


480.   1.    Solve  the  equations 


x'-^i/-{.z'  =  U.  (1) 

2x-3y-{-z  =  ll,  (2) 

x-\-2y-z=-6,  (3) 

Add  (2)  and  (3),  Sx-y  =  5;  ov,y  =  Sx-5.         (4) 

By  (3),       z  =  x-\-2y  +  6  =  x  +  6x-10-h6  =  7x-4..         (5) 
Substitute  values  of  y  and  z  in  (1), 

a^_|.9ic2_30a;  +  25  +  49a^-56a;4-16  =  14. 
Or,  59x'-S6x=-2T. 

ByM50,      .  =  M±V1^EI13  =  43^16  =  1  or  |I. 

oy  oy  oy 

Then  by  (4),  ^  =  3-5  or  §1-5= -2  01--^.    ' 
And  by  (5),   .  =  7-4  or  ^-4  =  3  or-|. 

Uy-b)(z-c)  =  a\  (1) 

2.  Solve  the  equations  i  (z—  c)  (x  —  a)  =  b\  (2) 

[(x-a)(y-b)  =  c',  (3) 

Multiply  (1),  (2),  and  (3), 

(x  -  of  ijj  -  bf  (z  -  cf  =  a'b^cK 
Whence,  (x  —  a)  (y  —  b)  (z  —c)  =  ±  abc.  (4) 

Divide(4)  by  (1),        x-a=  ±-,  ox  x  =  ^^^^. 

Divide  (4)  by  (2),        y-b  =  ±j,ovy  =  ^^. 

Divide  (4)  by  (3),         z—c=±  — ,  ov  z  =  -^ — . 

c  c 

(x{x^-y-\-z)  =  a\  (1) 

3.  Solve   the   equations    \  y(x -{-y -^z)  =  b^.  (2) 

[z(x-{-yi-z)  =  c'.  (3) 


SIMULTANEOUS   QUADRATIC   EQUATIONS         315 

Add  (1),  (2),  and  (3),     (x +  y  +  zf  =  a^ ^W  +  c\ 

(4) 


Then, 

x-^y-\-z=^±-yJa'  +  h''  +  c\ 

Divide  (1)  by  (4), 

X  — 1 

Va^  4-  6'  4-  c^ 

Divide  (2)  by  (4), 

w  =  4- 

Va^  4-  6^  +  c^ 

Divide  (3)  by  (4), 

.-1 

Va^  +  W  +  c^ 

EXERCISE  68 

(The  note  on  page  313  applies  with  equal  force  to  the   following 
examples.) 


Solve  the  following  : 

xy  =  a2&2. 

'xy  +  xz  =  lS-  x2. 

yz  =—  ab^. 
zx  =  —  a^b. 

7. 

-  2JZ +  yx  =  27  -  2/2. 
^zx  +  zy  =  S6-  z\ 

2. 

xh/z  =  12. 
xy'^z  =  6. 
xyz"^  =  18. 
xy  -^  yz  =  —  2>. 

8. 

(      xy  +  yz  —  zx  =  b. 

xy  —  yz  +  zx  =  c. 

_  -  xy  +  yz  -\-  zx  =  a. 

3. 

yz  +  zx  =  -S. 

-2x2  4-2/2-02  =  43. 

x-Sy  +  z  =  17. 

[     x  +  y-Sz  =  lS. 

izx-\-xy=-  35. 

9. 

(x  +  y){x  +  z)  = 

14. 

4. 
6. 

iM  +  z)(y  +  x)  = 

Xz  +  x){z-\-y)  = 

Sx"^  —  xy  -  xz  = 

bx-2y  = 

2. 

7. 

:4. 

=  1. 

10. 

x  +  y  -^  z  =  l2. 
xy  +  yz  +  zx  =  47. 

X2  +  2/2  -  02  _  0. 

4x4-3;?  = 

:-5. 

■  6  X  +  6  2/  =  5  xyz. 
-  3  2/  +  3  0  =  2  xyz. 
L  2  0  4-  2  X  =  xyz. 

[l  +  l-Uis. 

xyz 

11. 

6.   ^ 

1-1=1. 

y     X 

x2  4- 2/2 +  02  =  110. 

1  +  1  =  0. 
xy     z 

12. 

X  4-  y  -  0  =  4. 

X0  +  ^0  =  77. 

816  ADVANCED   COURSE  IN   ALGEBRA 

PROBLEMS  INVOLVING  SIMULTANEOUS  EQUATIONS  OF  A 
HIGHER  DEGREE  THAN  THE  FIRST 

481.  In  solving  problems  which  involve  simultaneous  equa- 
tions of  a  higher  degree  tfian  the  first,  there  will  usually  be 
more  than  one  set  of  values  of  the  unknown  numbers ;  only 
those  values  should  be  retained  which  satisfy  the  conditions  of 
the  problem. 

The  considerations  of  §§  261,  262,  and  463,  hold  for  simul- 
taneous equations  of  any  degree. 

EXERCISE  69 

1.  The  product  of  the  sum  of  two  numbers  by  the  smaller  is  21,  and 
the  product  of  their  difference  by  the  greater  is  4.     Find  the  numbers. 

2.  The  difference  of  the  squares  of  two  numbers  is  260  :  and  the  sum 

13 
of  the  numbers  is  —  their  difference.     Find  the  numbers. 
5 

3.  The  sum  of  the  squares  of  two  numbers  is  61,  and  the  product  of 
their  squares  is  900.    Find  the  numbers. 

4.  The  difference  of  the  cubes  of  two  numbers  is  316  ;  and  if  the  prod- 
uct of  the  numbers  be  added  to  the  sum  of  their  squares,  the  sum  is  79. 
Find  the  numbers. 

5.  Two  numbers  are  expressed  by  the  same  two  digits  in  reverse  order. 
The  sum  of  the  numbers  equals  the  square  of  the  sum  of  the  digits,  and 
the  difference  of  the  numbers  equals  5  times  the  square  of  the  smaller 
digit.    Find  the  numbers. 

6.  A  party  at  a  hotel  spent  a  certain  sum.  Had  there  been  five  more, 
and  each  had  spent  fifty  cents  less,  the  bill  would  have  been  $24.75.  Had 
there  been  three  less,  and  each  had  spent  fifty  cents  more,  the  bill  would 
have  been  $9.75.     How  many  were  there,  and  what  did  each  spend  ? 

7.  The  square  'of  "the  sum  of  two  numbers  exceeds  their  product  by 
84  and  the  sum  of  the  numbers,  plus  the  square  root  of  their  product, 
equals  14.    Find  the  numbers. 

8.  The  difference  of  the  cubes  of  two  numbers  is  728 ;  and  if  the  prod- 
uct of  the  numbers  be  multiplied  by  their  difference,  the  result  is  72. 
Find  the  numbers. 

9.  If  $  700  be  put  at  simple  interest  for  a  certain  number  of  years,  at 
a  certain  rate,  it  amounts  to  $883.75.  If  the  time  were  4  years  less,  and 
the  rate  1^  per  cent  more,  the  amount  would  be  $810.25.  Find  the  time 
and  the  rate. 


SIMULTANEOUS   QUADRATIC   EQUATIONS         317 

10.  If  the  digits  of  a  number  of  two  figures  be  inverted,  the  quotient 

4 
of  this  number  by  the  given  number  is  -,  and  their  product  2268.    Find 

the  number. 

11.  The  square  of  the  smaller  of  two  numbers,  added  to  twice  their 
product,  gives  7  times  the  smaller  number.  And  the  square  of  the  greater 
exceeds  the  product  of  the  numbers  by  6  times  the  smaller  number.  Find 
the  numbers. 

12.  A  and  B  travel  from  P  to  Q,  14  miles,  at  uniform  rates,  B  taking 
20  minutes  longer  than  A  to  perform  the  journey.  On  the  return,  each 
travels  one  mile  an  hour  faster,  and  B  now  takes  15  minutes  longer  than 
A.     Find  the  rates  of  travelling. 

13.  A  and  B  run  a  race  around  a  course  two  miles  long,  B  winning  by 
two  minutes.  A  now  increases  his  speed  by  two  miles  an  hour,  and  B 
diminishes  his  by  the  same  amount,  and  A  wins  by  two  minutes.  Find 
their  original  rates. 

14.  A  man  ascends  the  last  half  of  a  mountain  at  a  rate  one-half  mile 
an  hour  less  than  his  rate  during  the  first  half,  and  reaches  the  top  in  3f 
hours.  On  the  descent,  his  rate  is  one  mile  an  hour  greater  than  during 
the  first  half  of  the  ascent,  and  he  accomplishes  it  in  2f  hours.  Find  the 
distance  to  the  top,  and  his  rate  during  the  first  half  of  the  ascent. 

15.  The  square  of  the  second  digit  of  a  number  of  three  digits  exceeds 
twice  the  sum  of  the  first  and  third  by  3.  The  sum  of  the  first  and  second 
digits  exceeds  4  times  the  third  by  1.  And  if  495  be  subtracted  from  the 
number,  the  digits  will  be  inverted.     Find  the  number. 

16.  A  rectangular  piece  of  cloth,  when  wet,  shrinks  -  in  its  length, 
and  —  in  its  width.  If  the  area  is  diminished  by  lOf  square  feet,  and 
the  length  of  the  four  sides  by  5J  feet,  what  were  the  original  dimensions  ? 

17.  A  ship  has  provisions  for  36  days.  If  the  crew  were  16  greater,  and 
the  daily  ration  one-half  pound  less,  the  provisions  would  last  30  days.  If 
the  crew  were  2  less,  and  the  daily  ration  one  pound  greater,  they  would 
last  24  days.    Find  the  number  of  men,  and  the  daily  ration. 

18.  The  sum  of  two  numbers  is  5',  and  the  sum  of  their  fifth  powers  is 
1025.     Find  the  numbers. 

19.  A  man  lends  $2100  in  two  amounts,  at  different  rates  of  interest, 
and  the  two  sums  produce  equal  returns.  If  the  first  portion  had  been 
loaned  at  the  second  rate,  it  would  have  produced  .^48  ;  and  if  the  second 
portion  had  been  loaned  at  the  first  rate,  it  would  have  produced  .^27. 
Find  the  rates. 


318 


ADVANCED   COURSE   IN   ALGEBRA 


20.  A  can  do  a  piece  of  work  in  two  hours  less  time  than  B  ;  and 
together  they  can  do  the  work  in  1^  hours  less  time  than  A  alone.  How 
long  does  each  alone  take  to  do  the  work  ? 

21.  A  starts  to  travel  from  P  to  §,  and  at  the  same  time  B  starts  to 
travel  from  Q  to  P,  both  travelling  at  constant  rates.  A  reaches  ^  in  8 
hours,  and  B  reaches  P  in  18  hours,  after  they  have  met  on  the  road. 
How  many  hours  does  each  take  to  perform  the  journey  ? 

22.  A  and  B  travel  from  P  to  ^  and  back.  A  starts  one  hour  after  B, 
overtakes  him  at  a  point  two  miles  from  Q,  meets  him  32  minutes  after- 
wards, and  reaches  P  \\  hours  before  B.  Find  the  distance  from  P  to  §, 
and  the  rates  of  travel  of  A  and  B. 


GRAPHICAL  REPRESENTATION  OF  SIMULTANEOUS  QUAD- 
RATIC EQUATIONS  WITH  TWO  UNKNOWN    NUMBERS 

482.   1.   Consider  the  equation  2/^  =  4  cc  -f  4. 
We  have  y  =  ±2 Vx  + 1. 
If  0^  =  0,  y  =  ±2.     (A,B) 
If  x  =  l,  y  =  ±2V2.     (C,D) 
If  x  =  -l,  y  =  0.     (E) 
etc. 

For  any  positive  vahie  of  x^  or  for  any- 
negative  value  between  0  and  —l,y  has 
two  values ;  the  graph  extends  to  an  indefi- 
nitely great  distance  to  the  right  of  0. 

For  any  negative  value  of  x<  —1,  y  is  imaginary;  then,  no 
part  of  the  graph  lies  to  the  left  of  a  perpendicular  to  XX'  at  E. 

2.    Consider  the  equation  x^  ■ 
x^-l 


Here,  2/ 


or 


y 


r- 


2  \       JA'  0 

lfx  =  ±l,y=.0.     {A,  A') 

If  X  is  between  1  and  —1, 
y  is  imaginary ;  then,  no  part  of  the  graph  lies  between  per- 
pendiculars to  XX'  at  A  and  A\ 


SIMULTANEOUS  QUADRATIC   EQUATIONS         319 


If  x  =  ±2,  2/  =  ±^.     iB,C,B',a) 


For  any  positive  value  of  a;  >  1,  or  any  negative  value  <  —  1, 
y  has  two  values ;  then,  the  graph  has  two  branches,  each  of 
which  extends  to  an  indefinitely  great  distance  from  0. 

3.    Consider  the  equation  x^  -\-  y^  —  4:  x  -{■  2  y  =A. 
In  this  case,  it  is  convenient  to  first  locate  the  points  where 
the  graph  intersects  the  axes. 

If  y  =  0,      X^  —  4:X  =  4:j 

and  a;  =  2  ±  V8.     {A,  B) 

If  ic  =  0,   /  +  22/  =  4, 

and  2/  =  -l±V5.     {C,  D) 

We  may  write  the  given  equation 

or,  {y  +  iy  =  (l  +  x){^-x). 

If  x  =  -l  or  5,  2/  +  l  =  0,  and  y  =  -l.     (E,  F) 

If  x  has  any  positive  value  >  5,  or  any  negative  value  <  —  1, 

(1  -\-x)(5  —  x)  is  negative,  and  y  +  1  imaginary ;  then,  no  part 

of  the  graph  extends  to  the  right  of  F,  or  to  the  left  of  E. 
Again,  we  may  write  the  given  equation 

ar^_4a;  +  4  =  8-22/-2/^  or,  (x-2y  =  (4.-\-y){2-y). 
If  2/  =  -4  or  2,  iB-2  =  0,  and  x  =  2.     {O,  H) 
If  y  has  any  positive  value  >  2,  or  any  negative  value  <  —  4, 

(4  +  2/)  (2  —  y)  is  negative ;  then,  no  part  of  the  graph  extends 

above  H,  or  below  G. 

It  is  shown,  in  works  on  Analytic  Geometry,  that  the  graph  of  any  equa- 
tion of  the  second  degree,  with  two  unknown  numbers,  is  one  of  the  conic 
sections^  so-called  from  being  the  sections  of  a  cone  made  by  a  plane ;  either 
a  circle^  a  parabola,  an  ellipse,  a  hyperbola,  or  a  pair  of  straight  lines. 

The  graph  of  Ex.  1  is  a  parabola,  as  also  is  the  graph  of  any  equation 
of  the  form  y'^  =  ax,  or  y"^  =  ax  +  b. 

(The  graphs  of  §§  465  and  467  are  parabolas.) 

The  graph  of  Ex.  2  is  a  hyperbola,  as  also  is  the  graph  of  any  equation 
of  the  form  aoc^  —  by^  =  c,  it  a  and  b  are  numbers  of  like  sign. 

(The  hyperbola  has  two  branches.  The  graph  of  any  equation  of  the 
form  xy  =  a  is  a  hyperbola. ) 


320 


ADVANCED   COURSE  IN  ALGEBRA 


The  graph  of  Ex.  3  is  a  circle  ;  as  also  is  the  graph  of  any  equation  of 
the  form  x^  -}- y'^  =  a,  or  x^  -\-  y"^  -{-  ax  -{■  by  -\-  c  =  0. 

The  graph  of  any  equation  of  the  form  ax'^  +  by"^  =  c,  where  a,  6,  and  c 
are  numbers  of  like  sign,  and  a  and  b  unequal,  is  an  ellipse. 

(The  graph  of  any  equation  of  the  form  a^x^  —  b^y"^  =  0  is  the  pair  of 

straight  lines  whose  equations  are  y 


±f.) 


1.   Consider  the  equations 


483.  Graphical  Representation  of  Solutions  of  Systems  of  Simul- 
taneous Quadratic  Equations. 

I         2/2=4  a;. 

The  graph  of  2/^=4  a;  is  the  parabola  AOB. 

The  graph  of  Sx  —  y  =  5  is  the  straight 
line  AB,  intersecting  the  parabola  at  the 
points  A  Sind  B,  respectively. 

To  find  the  co-ordinates  of  A  and  B,  we 
solve  the  given  equations  (§  277) ;  the  solu- 
tion is  £c  =  1,  y  =  —  2,  and  x  —  — ,  y  =  — - 

y         o 

It  may  be  verified  in  the  figure  that  these  are  the  co-ordinates 
of  A  and  B,  respectively. 

Hence  (compare  §  277),  if  any  two  graphs  intersect,  the  co-ordi- 
nates of  any  point  of  intersection  form  a  solution  of  the  system  of 
equations  represented  by  the  graphs. 

2.   Consider  the  equations 

(x^-\-f  =  17. 

The  graph  of  x^  -\-y^=:  17  is  the 
circle  AD,  whose  centre  is  at  0,  and 
whose  radius  is  Vl7. 

The  graph  of  xy  =  4:  is  a  hyper- 
bola,  having   its   branches   in    the 
angles   XOY  and   X'OY',   respec- 
tively, and  intersecting  the  circle  at  the  points  A  and  B  in 
angle  XOY,  and  at  the  points  C  and  D  in  angle  X'OY', 


^m 


SIMULTANEOUS  QUADRATIC   EQUATIONS         321 


The  solution  of  the  given  equations  is 

a;  =  4,^=1 5  x=l,  y=4:;  x=—l,y  =  —4:;  and  a;=  — 4,  2/=  — 1. 

It  may  be  verified  in  the  figure  that  these  are  the  co-ordinates 
of  A,  B,  Cj  and  D,  respectively. 

3.   Consider  the  equations 

r     a^  +  42/'  =  4. 
[2x  +Sy  =-5. 

The  graph  of  ic^  +  4  2/^  =  4  is  the 
ellipse  AB,  intersecting  XX'  at 
points  2  to  the  right  and  2  to  the 
left  of  0,  and  YY'  at  points  1  above  and  1  below  0. 

The  graph  of2x-{-Sy  =  —  5  is  the  straight  line  CD. 

Substituting  x  = -^  "^      in  cc^  +  4  2/^  —  4,  we  obtain  the 

equation  25  2/^  +  30  2/  +  9  =  0,  which  has  equal  roots. 

9 


+  5 


Thus,  y 


■;  and  x  =  — 


5'  2  5 

The  equal  roots  signify  that  the  two  points  of  intersection 
coincide,  and  the  line  is  therefore  tangent  to  the  ellipse. 

In  general,  if  the  equation  obtained  by  eliminating  one  of 
the  unknown  numbers  has  equal  roots,  the  graphs  are  tangent 
to  each  other. 

4.   Consider  the  equations 
r9aj2-    2/'  =  -9. 
1    x-2y=-2. 

The  graph  of  9  o^  —  2/^  =  —  9  is  a  hyper- 
bola, having  its  branches  above  and  below 
0,  respectively. 

The  graph ofa;— 22/=— 2is  the  straight 
line  AB. 

Substituting  cc=2  2/— 2  in  9  a;^— 2/^=  —  9, 
we  obtain  the  equation  35  2/^ — 72  2/ + 45 = 0, 
which  has  complex  roots. 


322  ADVANCED   COURSE   IN   ALGEBRA 

Then,  the  graphs  have  no  point  of  intersection. 

In  general,  if  the  equation  obtained  by  eliminating  one  of  the 
unknown  numbers  has  no  real  root,  the  graphs  do  not  intersect 
each  other. 

EXERCISE  70 

Find  the  graphs  of  the  following  : 

1.  xy  =-  6.  3.    x2  -f  y2  _  4.  5.    9  x2  -  4  2/2  =  0. 

2.  x2  =  3  2/.  4.    2/^  =  5x-l.  6.    4x2  +  9  2/2  =  36. 

7.   4  x2  -  2/2  =  -  4.  8.    x2  +  y2  +  6  X  -  2  2/  =  15. 

Find  the  graphs  of  the  following  systems,  and  in  each  case  verify  the 
principles  of  §  483  : 


9. 

r  x2  +  4  2/2  =  4. 
ix   -     2/  =l• 

12. 

'    x2-     2/2  =  9. 
.5a;  _4y  ^_9. 

15. 

|x2  +  2/2  =  29. 
I         X2/=10. 

10. 

|x2+2/2=16. 
I          2/"'  =  6  X. 

13. 

'  2  x2  -  3  2/  =  5. 
,5x  +6y=-12. 

16. 

r  2x2+5  2/2=53. 
I  3x2-4  2/2= -24. 

11 

fy2_Sx=-S. 

I  x  +  22/  =  -2. 

14. 

'  9  x2  +  2/2  =  148. 
X2/  =  -  8. 

17. 

'ic2+2/H3x  =22. 
-     4x2-92/2=0. 

XX. 

INDETERMINATE  LINEAR  EQUATIONS  323 


XXII.    INDETERMINATE    LINEAR 
EQUATIONS 

484.  It  was  shown  in  §  269  that  a  system  of  m  independent 
linear  equations  containing  more  than  m  unknown  numbers, 
has  an  indefinitely  great  number  of  solutions. 

Such  a  system  is  called  indeterminate  (§  266). 
If,  however,  the  unknown  numbers  are  required  to  satisfy 
other  conditions,  the  number  of  solutions  may  be  finite. 

485.  Solution  of  Indeterminate  Linear  Equations  in  Positive 
Integers. 

We  shall  consider  in  the  present  chapter  the  solution  of 
indeterminate  linear  equations,  in  which  the  unknown  num- 
bers are  restricted  to  positive  integral  values. 

1.    Solve  7 x-\-By  —  118  in  positive  integers. 
Dividing  by  5,  the  smaller  of  the  two  coefficients, 

x-\-  —  -{-y  =  23-\--;  or,  — ^ —  =  2S-x-y. 

Since,  by  the  conditions  of  the  problem,  x  and  y  must  be 

2x 3 

positive  integers, must  be  an  integer. 

5 

Let  this  integer  be  represented  by  p. 

Then,       ^^~^  =p,ot2x-S  =  5  p.  (1) 

Dividing  (1)  by  2, 

a;-l-i  =  2p+|;  ov,x-l-2p=^^ 

Since  x  and  p  are  integers,  x  —  l  —  2p  is  an  integer ;  and 

therefore  ^-~—  must  be  an  integer. 

z 

Let  this  integer  be  represented  by  q. 

Then,  £±l  =  g.  oy^p^2q-\. 


324  ADVANCED  COURSE  IN   ALGEBRA 

Substituting  in  (1),    2x-3  =  10q-5. 

Whence,  x  =  5q  —  l.  (2) 

Substituting  this  value  in  the  given  equation, 

35  q -7 -\-5y  =  11S',  OT,y  =  25-7q.  (3) 

Equations  (2)  and  (3)  form  the  general  solution  in  iyitegers  of 
the  given  equation. 

By  giving  to  q  the  value  zero,  or  any  positive  or  negative 
integer,  we  shall  obtain  sets  of  integral  values  of  x  and  y  which 
satisfy  the  given  equation. 

If  q  is  zero,  or  any  negative  integer,  x  will  be  negative. 

If  q  is  any  positive  integer  >  S,  y  will  be  negative. 

Hence,  the  only  positive  integral  values  of  x  and  y  which 
satisfy  the  given  equation  are  those  obtained  from  the  values 
1,  2,  3  of  q. 

That  is,  a;  =  4,  2/  =  18 ;  a;  =  9,  2/  =  11 ;  and  ic  =  14,  ?/  =  4. 

2.    Solve  ^x  —  X3y  =  100  in  positive  integers. 
Dividing  by  8,  the  coefficient  of  smaller  absolute  value, 

x-y-^  =  12  +  ~;  0T,x-y-12  =  -^^—. 

Then,     ^ J'     must  be  an  integer. 
8 

Multiplying  by  5,  — ^^      must  also  be  an  integer. 
8 

Then,  Sy -\-^-{-2-\--  must  be  an  integer,  and  hence  ^"^ 
8  8  8 

must  be  an  integer ;  let  this  be  represented  by  p. 

Then,  ^"^    =p,  ot  y^Sp  —  4:. 

8 

Substituting  in  the  given  equation, 

8  X  -  104p  +  52  =  100,  or  a;  =  13p  4-  6. 

In  this  case  p  may  be  any  positive  integer. 

li  p  =  l,  x  =  19  and  ?/  =  4 ;  if  p  ==  2,  a.*  =  32  and  2/  =  12 ;  etc. 

Thus,  the  number  of  solutions  is  indefinitely  great. 


INDETERMINATE  LINEAR  EQUATIONS  325 

The  artifice  of  multiplying     ^  "^     by  5  saves  much  work  in  Ex.  2. 

8 

The  rule  in  any  case  is  to  multiply  the  numerator  of  the  fraction  by 
such  a  number  that  the  coefficient  of  the  unknown  quantity  shall  exceed 
some  multiple  of  the  denominator  by  unity. 

If  this  had  not  been  done,  the  last  part  of  the  solution  would  have 
stood  as  follows  : 

Let  ^1^  =p,  or  by  +  ^  =  8p.  (1) 

o 

Divide  hj  6,  y  +  -  =  p  -{- -^ ;  then     ^~     must  be  an  integer. ' 
5  5  6 

.     Let  Sp-4:  ^  ^^  or  3  j9  -  4  =  5  g.  (2) 

5 

Divide  hy3,;?-l--  =  g  +  ^;  then  ^^"^     must  be  an  integer. 

Let  2gjfl  =  r,  or  2q  +  l=3r.  (3) 

3 

Divide  by  2,  q-\-- =  r-\--;  then  must  be  an  integer. 

2  2  2 

r  —  1 

Let =  =  s,  or  r  =  2s  +  l. 

Substituting  in  (3),  2  g  +  1  =  6  s  +  3,  or  g  =  3  s  +  1. 

Substituting  in  (2),  3^)  —  4  =  15s-f5,  orj?  =  5s  +  3. 

Substituting  in  (1) ,  5  ?/  +  4  =  40  s  +  24,  or  ?/  =  8  s  +  4. 
Substituting  in  the  given  equation, 

8x- 104s -52  =  100,  oric  =  13a  +  19. 

These  values  of  %  and  y  differ  in  form  from  those  obtained  above  ;  but 
it  is  to  be  observed  that  13  s  +  19  and  8  s  +  4,  for  the  values  0,  1,2,  etc., 
of  s,  give  rise  to  the  same  series  of  positive  integers  as  \Zp  +  6  and  8  j?  —  4 
for  the  values  1,  2,  3,  etc.,  of  p. 

We  will  now  show  how  to  solve  in  positive  integers  two 
equations  involving  three  unknown  numbers. 

3.  In  how  many  ways  can  the  sum  of  $  14.40  be  paid  with 
dollars,  half-dollars,  and  dimes,  the  number  of  dimes  being 
equal  to  the  number  of  dollars  and  half-dollars  together  ? 

Let  X  —  number  of  dollars, 

y  —  number  of  half-dollars, 

and  » =*«  number  of  dimes. 


326  ADVANCED   COURSE   IN   ALGEBRA 

Then  by  the  conditions, 

(10x-^5y  +  z  =  144, 

1  x  +  y  =  z.  (1) 

Adding,  llx  +  6y-\-z  =  144  +  z, 

or,  11  a; +  6  2/ =  144.  (2) 

Dividing  by  6,    a;  +  ^  +  2/  =  24. 
o 

Then  —  must  be  an  integer ;  or,  x  must  be  a  multiple  of  6. 
6 

Let  x  =  6p,  where  p  is  an  integer. 

Substitute  in  (2),  66 p  +  6y  =  144,  or  2/  =  24  -  lip. 
Substitute  in  (1),  2;  =  6  p  +  24  - 1 1  j9  =  24  -  5 p. 

The  only  positive  integral  solutions  are  when  p  =  l  or  2. 

Therefore,  the  number  of  ways  is  two;  either  6  dollars,  13 
half-dollars,  and  19  dimes ;  or  12  dollars,  2  half-dollars,  and  14 
dimes. 

EXERCISE  71 

Solve  the  following  in  positive  integers : 

1.  Sx-\-6y  =  29.  6.  10x  +  7y  =  2Q7. 

2.  7  a; +  2  2/  =  39.  6.  23  a:  +  17  y  =  183. 

3.  6  a;  +  29  2/ =  274.  7.  8  x  + 71 2/  =  1933. 

4.  4  X  +  31  y  =  473.  8.  13  a;  +  50  ?/ =  1089. 

Solve  the  following  in  least  positive  integers  : 
9.   6x-7y  =  lS.  12.   8  a; -31 2/ =  10. 

10.  5  X  -  8  ?/  =  31.  13.    15  x  -  38  y  =  -  47. 

11.  14  X- 5?/ =  64.  14.   64x-19y  =  507. 

Solve  the  following  in  positive  integers  : 

ri2x+72/  +  25!  =  53.  r3x-32/-f-7;2  =  101. 

I    2x-lly  +  s=-25.  '   l4x-|-2y-3«  =  6. 

17.  In  how  many  different  ways  can  $  1.65  be  paid  with  quarter-dollars 
and  dimes  ? 

18.  In  how  many  different  ways  can  £2  Is.  be  paid  with  half-crowns, 
worth  2s.  6d.  each,  and  florins,  worth  2s.  each  ? 


INDETERMINATE   LINEAR  EQUATIONS  327 

19.  Find  two  fractions  whose  denominators  are  5  and  7,  respectively, 
whose  numerators  are  the  smallest  possible  positive  integers,  and  whose 

17 

difference  is  — . 
35 

20.  In  how  many  different  ways  can  $  7.15  be  paid  with  fifty-cent  pieces, 
twenty -five-cent  pieces,  and  twenty- cent  pieces,  so  that  twice  the  number 
of  fifty-cent  pieces,  plus  twice  the  number  of  twenty-cent  pieces,  shall 
exceed  the  number  of  twenty-five-cent  pieces  by  31  ? 

21.  A  farmer  purchased  a  certain  number  of  pigs,  sheep,  and  calves, 
for  .$138.  The  pigs  cost  |4  each,  the  sheep  $7  each,  and  the  calves  $9 
each  ;  and  the  whole  number  of  animals  purchased  was  23.  How  many 
of  each  did  he  buy  ? 

22.  In  how  many  different  ways  can  ^  5.45  be  paid  with  quarter-dol- 
lars, twenty-cent  pieces,  and  dimes,  so  that  twice  the  number  of  quarters, 
plus  5  times  the  number  of  twenty-cent  pieces,  shall  exceed  the  number 
of  dimes  by  36  ? 

486.  Every  linear  equation,  with  two  unknown  numbers,  x 
and  y,  can  be  reduced  to  one  of  the  forms 

ax±by=±c, 

where  a,  b,  and  c  are  positive  integers  which  have  no  common 
divisor. 

The  equation  ax  +  by=—c  cannot  be  solved  in  positive 
integers ;  for,  if  x,  y,  a,  and  b  are  positive  integers,  ax  -f-  by  must 
also  be  a  positive  integer. 

Again,  the  equations  ax±by  =  c  and  ax—by  =  —  c  cannot 
be  solved  in  positive  integers  if  a  and  b  have  a  common 
divisor. 

For,  if  X  and  y  are  positive  integers,  this  common  divisor 
must  also  be  a  divisor  of  ax  ±  by,  and  consequently  of  c ;  which 
is  contrary  to  the  hypothesis  that  a,  b,  and  c  have  no  common 
divisor. 


328  ADVANCED  COURSE  IN   ALGEBRA 


i 


XXIII.    RATIO  AND    PROPORTION 


\ 


487.  The  Ratio  of  one  number  a  to  another  number  &  is  the 
quotient  of  a  divided  by  b. 

Thus,  the  ratio  of  a  to  6  is  ^ ;  it  is  also  expressed  a:b. 

In  the  ratio  a:b,  a  is  called  the  first  term,  or  antecedent,  and 
b  the  second  term,  or  consequent. 

If  a  is  >  6,  the  ratio  a  :  &  is  called  a  ratio  of  greater  inequality;  if  a  is 
<  &,  it  is  called  a  ratio  of  less  inequality. 

The  ratio  of  the  product  of  the  antecedents  of  a  series  of  ratios  to  the 
product  of  the  consequents,  is  said  to  be  compounded  of  the  given  ratios. 

Thus,  ac  :  hd  is  compounded  of  the  ratios  a  :  h  and  c:  d. 

The  ratio  a^  :  6^  is  called  the  duplicate  ratio.,  the  ratio  a^  :  6^  the  tripli- 
cate ratio ^  and  the  ratio  Va  :  Vft  the  sub-duplicate  ratio.,  of  a  :  6. 

RATIO   OF   CONCRETE  MAGNITUDES 

488.  In  §  487.  we  considered  the  ratio  of  abstract  numbers 
only;  it  is,  however,  necessary  to  consider  the  ratio  of  two 
concrete  magnitudes  of  the  same  kind. 

If  a  concrete  magnitude  is  a  times  a  certain  unit,  and  another 
b  times  the  same  unit,  we  define  the  ratio  of  the  first  magnitude 
to  the  second  as  being  the  ratio  of  a  to  b. 

Thus,  the  ratio  of  two  lines  whose  lengths  are  2\  and  3|-  * 

inches,  respectively,  is  2\  -^  3J,  or  ■^- 

If  the  ratio  can  be  expressed  as  a  rational  number,  as  in  the 
above  illustration,  the  magnitudes  are  said  to  be  Commensurable. 

If  it  cannot  be  expressed  as  a  rational  number,  they  are  said 
to  be  Incommensurable. 


rjiCUi'UKTlUJN 

'  -    489.   A  Proportion  is  an  equation  whose  members  are  equal 


RATIO  AND  PROPORTION  329 


6  =  c  :  d,  or  -  =  -, 
b      d 


Thus,  if  a:  b  and  c :  d  are  equal  ratios, 

is  a  proportion. 

The  symbol  : :  is  sometimes  used  in  place  of  the  sign  of  equality  in  a 
proportion. 

490.   In  the  proportion  a:b  =  c:d,  a  is  called  the  first  term, 
b  the  second,  c  the  third,  and  d  the  fourth. 
-^    The  first  and  third  terms  of  a  proportion  are  called  the  ante- 
cedents, and  the  second  and  fourth  terms  the  consequents. 
_X    The  first  and  fourth  terms  are  called  the  extremes,  and  the 

second  and  third  terms  the  means. 
vy    If  the  means  of  a  proportion  are  equal,  either  mean  is  called 
a  Mean  Proportional  between  the  first  and  last  terms,  and  the 
last  term  is  called  a  Third  Proportional  to  the  first  and  second 
terms. 

Thus,  in  the  proportion  a:b  =  b:c,  b  is  a  mean  proportional 
between  a  and  c,  and  c  is  a  third  proportional  to  a  and  b. 

A  Fourth  Proportional  to  three  numbers  is  the  fourth  term  of 
a  proportion  whose  first  three  terms  are  the  three  numbers 
taken  in  their  order. 

Thus,  in  the  proportion  a:b  =  c:d,  d  is  a  fourth  proportional 
to  a,  b,  and  c. 

N    A  Continued  Proportion  is  a  series  of  equal  ratios,  in  which 
each  consequent  is  the  same  as  the  following  antecedent ;  as, 

a:b  =  b:c  =  c:d  =  d:e. 

PROPERTIES  OF  PROPORTIONS 

^  491.   In  any  proportion,  the  product  of  the  extremes  is  equal  to 
the  product  of  the  means. 

Let  the  proportion  be     a:b  =  c:d. 

Then  by  §489,  -  =  -. 

b     d 

Clearing  of  fractions,        ad  =  6c. 


330  ADVANCED  COURSE   IN   ALGEBRA 

492.   From  the  equation  ad  =  be  (§  491),  we  obtain 

be  J      ad         ad        -,    -,     be 
a  =  — ,  0  =  — ,  c  =  — ,  and  d  =  — 
deb  a 

That  is,  m  any  proportion,  either  extreme  equals  the  produet 
of  the  means  divided  by  the  other  extreme;  and  either  mean  equals 
the  produet  of  the  extremes  divided  by  the  other  mean. 

^  493.  (Converse  of  §  491.)  If  the  produet  of  two  numbers  be 
equal  to  the  produet  of  two  others,  one  pair  may  be  made  the 
extremes,  and  the  other  pair  the  means,  of  a  proportion. 

Let  ad  =  be.  (1) 

'    Dividing  by  bd,  -  =  -• 

b     d 

Then,  a  :  b  =  e  :  d,  ov  e  :  d  =  a  :  b. 

In  like  manner,  by  dividing  the  members  of  (1)  by  ab,  then 
by  cd,  and  then  by  ae,  we  have 

d:b  =  e:  a,  ov  c  :  a  =  d  .  b, 

a:  e  =  b  :  d,  ov  b:  d  =  a:c, 

and  d  :  e  =  b  :  a,  ov  b  :  a  =  d:  c. 

\  494.  In  any  proportion,  the  terms  are  in  proportion  by  alter- 
nation ;  that  is,  the  means  ean  be  interehanged. 

Let  the  proportion  be        a:  b  =  c:  d. 

Then  by  §  491,  ad  =  be.,  (1) 

Whence,  by  §  493,  a:e  =  b:d. 

We  also  have  from  (1),  by  §  493, 

d:b  =  c:a. 
That  is,  in  any  proportion,  the  extremes  can  he  interchanged. 

>»  495.  In  any  proportioyi,  the  terms  are  in  proportion  by  Inver- 
sion ;  that  is,  the  second  term  is  to  the  first  as  the  fourth  term  is 
to  the  third. 

Let  the  proportion  be         a  :  b  =  e  :  d. 

Then  by  §  491,  ad  =  be. 


RATIO   AND  PROPORTION  331 

Whence,  by  §  493,  h'.a  =  d'.c. 

It  follows  from  the  above  that,  in  any  proportion^  the  means  can  he 
written  as  the  extremes^  and  the  extremes  as  the  means. 

~<  496.  A  mean  proportionalJ)etween  two  numbers  is  equal  to  the 
square  root  of  their  product. 

Let  the  proportion  be      a:b  —  b  :  c. 

Then  by  §  491,       b^  =  ac,  and  b  =  Vac. 

1  497.  In  any  proportion,  the  terms  are  in  proportion  by  Com- 
position; that  is,  the  sum  of  the  first  two  terms  is  to  the  first 
term  as  the  sum  of  the  last  two  terms  is  to  the  third  term. 

Let  the  proportion  be     a  :  b  =  c  :  d. 

Then,  ad  =  be. 

Adding  each  member  of  the  equation  to  ac, 

ac  -j-  ad  =  ac-\-  be,  or  a{c  -\-  d)  =  c(a  +  5). 
Then  by  §  493,         a-\-b\  a  =  c -{- d:c. 
We  may  also  prove  a  -\- b  :h  ■=  c  ■{■  d:  d. 


4 


498.  In  any  proportion^  the  terms  are  in  proportion  by  Division ; 
that  is,  the  difference  of  the  first  two  terms  is  to  the  first  term  as 
the  difference  of  the  last  two  terms  is  to  the  third  term. 

Let  the  proportion  be     a:b  =  c:d. 

Then,  ad  —  be. 

Subtracting  each  member  of  the  equation  from  ac, 

ac  —  ad  =  ac  —  be,  or  a(c  —  d)  =  c(a  —  b). 

Then,  a  —  b:a  =  c  —  d:c. 

We  may  also  prove         a  —  b:b  =  c  —  did. 

499.  In  any  proportion,  the  terms  are  in  proportion  by  Com- 
position and  Division ;  that  is,  the  sum  of  the  first  two  terms  is 
to  their  difference  as  the  sum  of  the  la^t  two  terms  is  to  their 
difference. 

Let  the  proportion  be     a:b  =  c:d 


832  ADVANCED  COURSE  IN  ALGEBRA 


Then  by  §497,  a±h^c_±d^ 

a            c  ^ 

And  by  §  498,               ^—^  =  ^^.  (2) 

a            c  ^ 


Dividing  (1)  by  (2), 


a  +  b     c-{-d 


a  —  b     c  —  d 
Whence,  a-\-b:a  —  b  =  c-{-d:G  —  d. 

500.  In  any  proportion,  if  the  first  two  terms  be  multiplied  by 
any  number,  as  also  the  last  two,  the  resulting  numbers  will  be  in 
proportion. 

Let  the  proportion  be     a'.b  =  c:d. 

Then,  »  =  «;  and  hence  ??^  =  !^. 
b     d  mb     nd 

Therefore,  ma  :  mb  =  nc  :  nd. 

We  may  also  prove  ^ :  A  =  ^  .  ^. 

m    m     n    n 

(Either  m  or  n  may  be  unity  ;  that  is,  the  terms  of  either  ratio  may  be 
multiplied  or  divided  without  multiplying  or  dividing  tlje  terms  of  the 
other.)  ,^,  ,      _, 

501.  In  any  proportion,  if  the  first  and  third  terms  be  multi- 
plied by  any  number,  as  also  the  second  and  fourth  terms,  the 
resulting  numbers  loill  be  in  proportion. 

Let  the  proportion  be     a:b  =  c:d. 

Then,  ^  =  ^;  and  hence  ™!  =  ™«. 
b      d  nb      nd 

Therefore,  ma  :nb  =  mG:  nd. 

trr  1  Ct        b  C        d 

We  may  also  prove  _:_  =  —  :  — 

m    n     m    n 

(Either  w  or  w  may  be  unity. ) 

^    502.   In  any  number  of  proportions,  the  products  of  the  cor- 
responding terms  are  in  proportion. 

Let  the  proportions  be  a  :  b  =  c  :  d, 
and  e:f=g:h. 


\ 


RATIO  AND  PROPORTION  333 

Then,  ^  =  '    and^  =  f. 

'  b      d  f     h 

,r  ,,.  T   .  a     e       c     q         ae      cq 

Multiplying,  _x_^  =  -x2,o.--  =  X 

Whence,  ae:hf=cg:  dh. 

In  like  manner,  the  theorem  may  be  proved  for  any  number 
of  proportions. 

503.  The  quotients  of  the  corresponding  terms  of  two  pro- 
portions are  in  proportion. 

Let  the  proportions  be  a  :  6  =  c :  c?, 

and  e:f=g:h. 

Then,  -=-,  and  :£  =  -• 

b      d  eg 

TT71,  Ci      f      c^,h  abed 

Whence,  -x=^  =  -X-;  or,  --f--  =  --f--. 

b      e      d     g  e      f     g     h 

Then,  «:A  =  f:f. 

e    f     g    h 

504.  In  any  proportion,  like  powers  or  like  principal  roots  of 
the  terms  are  in  proportion. 

Let  the  proportion  be     a:b=^  c:d. 

Then,  -  =  - :  and  hence    ^  =  ^. 
'  b      d'  b""     d"" 

Therefore,  a** :  ^'^  =  c"  :  d\ 

We  may  also  prove         \/a  :  Vb  =  Vc  :  Vd. 

^^505.  In  a  series  of  equal  ratios,  any  antecedent  is  to  its  con- 
sequent as  the  sum  of  all  the  antecedents  is  to  the  sum  of  all  the 
consequents. 

Let  a:b  =  c:d  =  e:f 

Then  by  §  491,  ad  =  be, 

and  af=  be. 

Also,  ab  =  ba. 


334       ADVANCED  COURSE  IN  ALGEBRA 

Adding,  a(b  -\-  d -i-f)  =  b(a -\-  c -^  e). 

Whence,  a:b  =  a -{- c -{- e:b -]- d-j-f.      (§493) 

In  like  manner,  the  theorem  may  be  proved  for  any  number 
of  equal  ratios. 

Q    506.    To  prove  that  if         -  =  -  =  !=..., 
^  b      d     f 

1 

then  each  of  these  equal  ratios  equals     {  ^      ~  ^ — IL JI — 

\pb^J^qd^-^rr+-- 

If  -  =  -  =  -1  =  ...  =  yfc,  then  a  =  bk,  c  =  dk,  e  =fk,  etc. 
b      d      e 

Then,    pa""  +  qc""  +  re"" -{-"-=  p  (bky  +  q  (dky  +  r  (fky  +  •  •  • 

=  A;"(p6'*  +  gd"  +  rf"  +  •••). 
Therefore,  k^  =  P^"  +  gc- +  re- +  ...^ 

Or  A;  =  /^i>«"  +  gc"  +  re-  +  --Y 

'  \pb''  +  qd''  +  rf''-{-  '"J 

Q  507.   7/*  ^/iree  numbers  are  in  continued  proportion,  the  first  is 
to  the  third  as  the  square  of  the  first  is  to  the  square  of  the  second. 

Let  the  proportion  be  a:b  =  b:c. 

Then,  ?  =  5. 

b     c 

Therefore,  ^  x  ^  =  ^X  ?,  or  «  =  ^. 

b      c     b      b  c      b^ 

Whence,  a:c  =  a^:W. 

\J  508.   If  four  numbers  are  in  continued  proportion,  the  first  is  to 
the  fourth  as  the  cube  of  the  first  is  to  the  cube  of  the  second. 

Let  the  proportion  be  a:b  =  b  \  c  =  c:d. 

rT^^  a         b         C 

Then^  -  =  -  =  -• 

"*  bed 

Therefore,  ^  x  ^  x  ^  =  f  X  ?  X  «   or  «  =  ^'. 

b      G      d      b      b      b  d     b" 

Whence,  a:d  =  a^:  ¥. 


RATIO   AND  PROPORTION  335 

509.  Examples. 

1.  If  x:y  =  {x-{-z)'^'.{y-\-zf,  prove  z  a  mean  proportional 
between  x  and  y. 

From  the  given  proportion,  by  §  491, 

y{x  -\-  zf  =  x(y  +  zy,  or  ib-y  +  2  xyz  +  yz^  =  xy^  +  2  ic?/^!  ■+-  xz^- 
Transposing,  x^y  —  xy^  =  xz^  —  2/2;l 

Dividing  hj  x  —  y^  xy  =  z\ 

Therefore,  2;  is  a  mean  proportional  between  x  and  y. 
The  theorem  of  §  499  saves  work  in  the  solution  of  a  certain 
class  of  fractional  equations. 

2.  Solve  the  eq uation  ^'  +  a; - 1  ^ ^-2 ^ 

ar  —  x  —  1     x-\-2 

By  composition  and  division, 

2a;^-2_  2x  ^ -\  _     x 

2x     --4'''''      X     ~     2 

Clearing  of  fractions,  2  af'  —  2  =  —  aj^ 

Then,  3  x^  =  2,  and  a;  =  ±-^-. 

3.  Prove  that  if  -  =  -,  then 

h     d 

a2  _  52 :  a2  -  3  ao  =  c2  -  ^2 :  c2  -  3  cd. 

Let  -  =  -  =  Xj  then,  a  =  6a;. 
6     d 


Therefore, 

c" 

1 

a'- 

-b' 

Wx' 

-b' 

x'- 

-1 

d? 

^     & 

-d^ 

a'- 

Sab 

6V- 

■3bh 

:     x'- 

-3  a; 

d^ 

3  c     c^- 
d 

-3c(i 

Then 

a^-ft^ 

a'- 

Sab  = 

c'- 

d^',& 

-3cd 

EXERCISE  72 

1.  Find  the  third  term  of  a  proportion  whose  first,  second,  and  fourth 

terms  are  - ,  - ,  and  - ,  respectively. 
4    6  9^ 

2.  Find  a  third  proportional  to  —  and  — 

y         12 


3S6  ADVANCED   COURSE   IN  ALGEBRA 

3.  Find  a  mean  proportional  between  l\l  and  24|. 

4.  Find  a  fourth  proportional  to  4|,  5|,  and  If. 

5.  Find  a  third  proportional  to  a^  +  27  and  a  +  3. 

^2 /v. 1 2  x^ 9  jC  4-  20 

6.  Find  a  mean  proportional  between  and  '- — — — . 

.    x-6  ic  +  3 

Solve  the  following  equations  : 

-     3a;-8_2a;-  5_  g     x^-\-2x-S_Sx  +  2 

'3a;  +  4~2x  +  7'  '    x^-2x-S     3ic-2* 


9     a;2  +  3x-l_a;g  +  2a;+l 
x^-Zx+1     x2-2x-l 


10. 


11. 


Va;2  +  1  +  V  a;2  -  1  _  Vgg  _  2  g  +  2  +  Va^  _  2a 
Vic2  +  l  -  Vx2-  1      \/a2  _  2  a  +  2  -  Va'^  -  2  a 

X  —  y     a  —  b 

x^  +  x  +  y'^  _a^-\-  a  +  b^ 


2  -  X  +  ?/2     a2  -  a  +  62 

12.  If  a  :  &  =  c  :  c?  and  e  :f  =  g  :h,  prove 

ae  +  bf-.ae  —  bf=  eg  +  dh  :cg  —  dh. 

13.  Find  two  numbers  such  that,  if  9  be  added  to  the  first,  and  7  sub- 
tracted from  the  second,  they  will  be  in  the  ratio  9:2;  while  if  9  be  sub- 
tracted from  the  first,  and  7  added  to  the  second,  they  will  be  in  the  ratio 
9:11. 

14.  Find  two  numbers  in  the  ratio  a  :  6,  such  that,  if  each  be  increased 
by  c,  they  shall  be  in  the  ratio  m:n. 

15.  Find  three  numbers  in  continued  proportion  whose  sum  is  28|, 

2 
such  that  the  quotient  of  the  first  by  the  second  shall  be  -• 

o 

16.  Find  a  number  such  that,  if  it  be  added  to  each  term  of  the  ratio 

8 : 5,  the  result  is  —  of  what  it  would  have  been  if  the  same  number 

20 
had  been  subtracted  from  each  term. 

17.  The  second  of  three  numbers  is  a  mean  proportional  between  the 
other  two.  The  third  number  exceeds  the  sum  of  the  other  two  by  20  ; 
and  the  sum  of  the  first  and  third  exceeds  three  times  the  second  by  4. 
Find  the  numbers. 

18.  If8a  —  56:7a  —  45  =  86-5c:76  —  4  c,  prove  c  a  third  propor- 
tional to  a  and  6. 

19.  lia  +  b  +  c  +  d:a  +  b  =  a  —  b  +  c  —  d:a  —  b,  prove 

a:b  =  C:d. 


RATIO   AND   PROPORTION  337 


20.  lix  +  y:y  +  z  =  y/x^  —  y'^  :  y/y'^  —  2^,  prove  y  a  mean  proportional 
between  x  and  z. 

21.  A  is  following  B  along  a  certain  road,  when  B  turns  and  walks  in 
the  opposite  direction;  if  A  and  B  approach  each  other  five  times  as  fast 
as 'before,  compare  their  rates. 

22.  If  4  silver  coins  and  11  copper  coins  are  worth  as  much  as  2  gold 
coins,  and  5  silver  coins  and  19  copper  coins  as  much  as  3  gold  coins,  find 
the  ratio  of  the  value  of  a  gold  coin,  and  the  value  of  a  silver  coin,  to  the 
value  of  a  copper  coin. 

23.  Given  2  (a^  +  ah)x  +  (a^  +  2  h'^)y  =  {a^  -  b^)x  +  (2  a^  +  b^)y,  find 
the  ratio  of  x  to  y. 

24.  Given  ?  +  ^  =  ^  +  ^  =  ^-f^,  fi^d  the  ratio  of  x  to  y,  and  of  x  to  z. 

b     a     c     a     c     b 

25.  If  -  =  -,  prove, 

b     d 

(a)   3  a2  -  4  a6  :  2  a6  +  7  62  =  3  c2  -  4  cci :  2  ctZ  +  7  (^2. 

lb)   a^  +  6  ab'^ :  a^b  -  bb^  =  c^  +  Q  cdC^  -.cH  -  b  #. 

26.  The  sum  of  four  numbers  in  proportion  is  32.  The  sum  of  the 
means  exceeds  the  sum  of  the  extremes  by  4  ;  and  the  sum  of  the  conse- 
quents exceeds  the  sum  of  the  antecedents  by  16.     Find  the  numbers. 

27.  A  passenger  observes  that  a  train  passes  him,  moving  in  the 
opposite  direction,  in  3  seconds ;  while,  if  it  had  been  moving  in  the  same 
direction,  it  would  have  passed  him  in  13  seconds.  Compare  the  rates  of 
the  trains. 

28.  Each  of  two  vessels  contains  a  mixture  of  wine  and  water.  A 
mixture  consisting  of  equal  measures  from  the  two  vessels  is  composed  of 
wine  and  water  in  the  ratio  3:4;  another  mixture  consisting  of  14  meas- 
ures from  the  first  and  21  measures  from  the  second,  is  composed  of  wine 
to  water  in  the  ratio  2  : 3.    Find  the  ratio  of  wine  to  water  in  each  vessel. 

29.  If  ^  =  ^  =  -^,  prove 

b     d     f 

(a)  a^  +  c"^  +  e^ :b^  +  d^  +  P  =  ac  +  ce  +  ea  :bd  -h  df  +  fb. 
(6)  a*  +  c*  +  c*  :  6*  +  d*  +/*  =(««  +  c2  +  e'^y  :  (b^  +  d^  +Py. 

30.  If  a  +  6,  6  +  c,  and  c  +  a  are  in  continued  proportion,  prove 

a  +  b:b  +  c  =  c  —  a:a  —  b. 

31.  Find  four  numbers  in  proportion  such  that  the  sum  of  the  means 
is  21,  and  of  the  extremes  39  ;  and  twice  the  last  term  exceeds  three  times 
the  sum  of  the  first  two  terms  by  30. 

32.  If  a,  6,  c,  and  d  are  in  continued  proportion,  prove 

3a  +  4d:2a-5d  =  3a8  +  4  68:2a3_6  63. 


338  ADVANCED   COURSE  IN   ALGEBRA 


XXIV.    VARIATION 

510.  One  variable  number  (§  245)  is  said  to  vary  directly  as 
another  when  the  ratio  of  any  two  values  of  the  first  equals 
the  ratio  of  the  corresponding  values  of  the  second. 

It  is  usual  to  omit  the  word  "directly,"  and  simply  say  that  one 
number  varies  as  another. 

Thus,  if  a  workman  receives  a  fixed  number  of  dollars  per 
diem,  the  number  of  dollars  received  in  m  days  will  be  to  the 
number  received  in  n  days  as  m  is  to  w. 

Then,  the  ratio  of  any  two  numbers  of  dollars  received  equals 
the  ratio  of  the  corresponding  numbers  of  days  worked. 

Hence,  the  number  of  dollars  which  the  workman  receives 
varies  as  the  number  of  days  during  which  he  works. 

The  symbol  cc  is  used  to  express  variation;  thus,  ace 6  is 
read  "  a  varies  as  6.'^ 

511.  One  variable  is  said  to  vary  inversely  as  another  when 
the  first  varies  directly  as  the  reciprocal  of  the  second. 

Thus,  the  number  of  hours  in  which  a  railway  train  will 
traverse  a  fixed  route  varies  inversely  as  the  speed ;  if  the 
speed  be  doubled,  the  train  will  traverse  its  route  in  oyie-half 
the  number  of  hours. 

One  variable  is  said  to  vary  as  two  others  jointly  when  it 
varies  directly  as  their  product. 

Thus,  the  number  of  dollars  received  by  a  workman  in  a 
certain  number  of  days  varies  jointly  as  the  number  which  he 
receives  in  one  day,  and  the  number  of  days  during  which  he 
works. 

One  variable  is  said  to  vary  directly  as  a  second  and  inversely 
as  a  third,  when  it  varies  jointly  as  the  second. and  the  recip- 
rocal of  the  third. 

Thus,  the  attraction  of  a  body  varies  directly  as  the  amount 
of  matter,  and  inversely  as  the  square  of  the  distance. 


VARIATION  339 

512.  IfxcxLy,  then  x  equals  y  multiplied  by  a  constant  number. 
Let  «'  and  2/'  denote  a  j^a;ed  pair  of  corresponding  values  of 

X  and  y,  and  x  and  y  any  other  pair. 

By  the  definition  of  §  510,  -  =  -7;  or,  x  =  -,y. 

a;' 
Denoting  the  constant  ratio  —  by  m,  we  have 

X  =  my. 

513.  It  follows  from  §§  511  and  512  that: 

1.  Ifx  varies  inversely  as  y,  x  =  — . 

2.  Ifx  varies  jointly  as  y  and  z,  x  —  myz. 

mil 

3.  Ifx  varies  directly  as  y  and  inversely  as  2;,  x  =  ^-- 

z 

The  converse  of  each  statement  of  §§  512  and  513  is  also  true  ;  that  is, 
if  X  equals  y  multiplied  by  a  constant,  xazy,  etc. 

514.  Ifx<x:y,  and  y  ccz,  then  x  ccz.  V.^  (^n^*>-^ 
By  §  512,  iix<xy,            x  =  my.  (1) 
And  if  2/  oc  2;,                      y  =  nz. 
Substituting  in  (1),           x  =  mriz. 

Whence,  by  §  513,  x  cc  z. 

515.  Ifxaoy  when  z  is  constant,  and  xccz  when  y  is  constant,    O-nn' 
then  xocyz  when  both  y  and  z  vary. 

Let  y'  and  z'  be  the  values  of  y  and  z,  respectively,  when  x 
has  the  value  x'. 

Let  y  be  changed  from  y'  to  y",  z  remaining  constantly  equal 
to  z',  and  let  x  be  changed  in  consequence  from  x'  to  X 

Then  by  §510,  -|=4  (1) 

Now  let  z  be  changed  from  z'  to  z",  y  remaining  constantly 
equal  to  y",  and  let  x  be  changed  in  consequence  from  X  to  x". 

Then,  ^=^.  (2) 


340  ADVANCED  COURSE   IN   ALGEBRA 

Multiplying (1) by. (2),   $,='$^,-  (3) 

Now  if  both  changes  are  made,  that  is,  y  from  y'  to  y"  and 
z  from  z'  to  2",  a;  is  changed  from  x'  to  x",  and  1/2;  is  changed 
from  y'z'  to  y"z". 

Then  by  (3),  the  ratio  of  any  two  values  of  x  equals  the 
ratio  of  the  corresponding  values  of  yz ;  and,  by  §  510,  x  oc  yz' 

In  like  manner  it  may  be  proved  that  if  there  are  any 
number  of  variables  x,  y,  z,  u,  etc.,  such  that  ic  x  ?/  when  z,  u,  etc., 
are  constant,  xccz  when  y,  u,  etc.,  are  constant,  etc.,  then  if  all 
the  variables  y,  z,  u,  etc.,  vary,  x  varies  as  their  product. 

The  following  is  an  illustration  of  the  above  theorem  : 

It  is  known,  by  Geometry,  that  the  area  of  a  triangle  varies  as  the  base 

when  the  altitude  is  constant,  and  as  the  altitude  when  the  base  is 

constant. 

Hence,  when  both  base  and  altitude  vary,  the  area  varies  as  their 

product. 

PROBLEMS 

516.  Problems  in  variation  are  readily  solved  by  converting 
the  variation  into  an  equation  by  aid  of  §§  512  or  513. 

1.  If  a;  varies  inversely  as  y,  and  equals  9  when  y  =  S,  find 
the  value  of  x  when  y  =  18. 

If  x  varies  inversely  2iS  y,  x  =  —  (§  513). 

y 

Putting  oJ  =  9  and  2/  =  8,   9  =  ^,  or  m  =  72. 

8 

72  72 

Then,  x  =  — ;  and,  if  2/  =  18,  a?  =  — -  =  4. 
y  18 

2.  Given  that  the  area  of  a  triangle  varies  jointly  as  its 
base  and  altitude,  what  will  be  the  base  of  a  triangle  whose 
altitude  is  12,  equivalent  to  the  sum  of  two  triangles  whose 
bases  are  10  and  6,  and  altitudes  3  and  9,  respectively  ? 

Let  B,  H,  and  A  denote  the  base,  altitude,  and  area,  respec- 
tively, of  any  triangle,  and  5'  the  base  of  the  required  triangle. 
Since  A  varies  jointly  as  B  and  H,  A  =  mBH  (§  513). 


VARIATION  341 

Then  the  area  of  the  first  triangle  is  m  x  10  x  3,  or  30  m, 
and  the  area  of  the  second  is  m  x  6  X  9,  or  54  m ;  whence,  the 
area  of  the  required  triangle  is  30  m  +  54  m,  or  84  m. 

But  the  area  of  the  required  triangle  is  also  m  x  B'  x  12. 

Therefore,  12  mB'  =  84  m,  and  B'  =  7. 

EXERCISE  73 

2  3 

1.  If  a;  varies  inversely  as  y,  and  equals  -  when  y  =  -i  what  is  the  value 

3  3  4 

of  V  when  x  =  ~? 
-^  2 

2.  Itycc  ^2,  and  equals  40  when  z  =  10,  what  is  the  value  of  y  in  terms 

of  02? 

2  3  4 

3.  If  z  varies  jointly  as  x  and  y,  and  equals  -  when  x  =  -  and  y  =  -. 

4  53  4  5 

what  is  the  value  of  0  when  x=  -  and  y  =  -? 
\  o  4 

Q 

4.  If  X  varies  directly  as  y  and  inversely  as  0,  and  is  equal  to  —  when  y 

16 
=  27  and  z  =  64,  what  is  the  value  of  x  when  y  =  9  and  2:  =  32  ? 

5.  If  5  X  +  8  X  6  y  —  1,  and  x  =  6  when  2/=  —3,  what  is  the  value  of  x 
when  y  =  7  ? 

6.  If  x'^  Qc  y^,  and  ic=4  when  y =4,  what  is  the  value  of  y  when  aj  =  -  ? 

7.  The  surface  of  a  cube  varies  as  the  square  of  its  edge.  If  the  sur- 
face of  a  cube  whose  edge  is  2^  inches  is  32§  square  inches,  what  will  be 
the  edge  of  a  cube  whose  surface  is  30|  square  inches  ? 

8.  The  distance  fallen  by  a  body  from  a  position  of  rest  varies  as  the 
square  of  the  time  during  which  it  falls.  If  a  body  falls  1029^  feet  in  8 
seconds,  how  long  will  it  take  to  fall  402^^^  feet  ? 

9.  If  7  men  in  4  weeks  earn  $238,  how  many  men  will  earn  $127.50 
in  3  weeks ;  it  being  given  that  the  amount  earned  varies  jointly  as  the 
number  of  men  and  the  number  of  weeks  during  which  they  work. 

10.  A  circular  plate  of  lead,  17  inches  in  diameter,  is  melted  and  formed 
into  three  circular  plates  of  the  same  thickness.  If  the  diameters  of  two 
of  the  plates  are  8  and  9  inches,  respectively,  find  the  diameter  of  the 
other ;  it  being  given  that  the  area  of  a  circle  varies  as  the  square  of  its 
diameter. 

11.  If  y  equals  the  sum  of  two  numbers  which  vary  directly  as  a;*  and 

29 
inversely  as  x,  respectively,  and  y—  —53  when  a;  =  —3,  and  y=  —  when  x 

1  2 

=  2,  what  is  the  value  of  y  when  x  =  -  ? 

2 


342  ADVANCED  COURSE  IN  ALGEBRA 

12.  If  X  equals  the  sum  of  two  numbers,  one  of  which  varies  directly 
as  ?/2  and  the  other  inversely  as  z^^  and  ic  =  45  when  y  =  \  and  ^  =  1,  and 
X  =  40  when  2/  =  2  and  5;  =  3,  find  the  value  of  y  when  x  =  87  and  ^  =  1. 

13.  If  y  equals  the  sum  of  three  numbers,  the  first  of  which  is  constant, 
and  the  second  and  third  vary  as  a:'-^  and  x^,  respectively,  and  ?/  =  —  50 
when  X  =  2,  30  when  x  =  —  2,  and  110  when  x  =9  —  3,  find  the  expression 
for  y  in  terras  of  x. 

14.  The  volume  of  a  circular  coin  varies  jointly  as  its  thickness  and 
the  square  of  the  radius  of  its  face.  Two  coins  whose  thicknesses  are 
5  and  7,  and  radii  of  faces  60  and  30,  respectively,  are  melted,  and 
formed  into  100  coins,  each  3  units  thick.  Find  the  radius  of  the  face  of 
the  new  coin. 

15.  The  distance  travelled  by  a  man,  in  any  hour  after  the  first,  equals 
a  constant  number  of  miles,  plus  a  number  of  miles  which  varies  inversely 
as  the  number  of  hours  travelled  before  that  hour.  If  he  travels  12  miles 
in  the  6th  hour,  and  8  in  the  11th,  how  far  does  he  travel  in  the  21st  hour  ? 

16.  If  the  weight  of  a  spherical  shell,  two  inches  thick,  is  —  of  its 

weight  if  solid,  find  its  diameter ;   it  being  given  that  the  volume  of  a 
sphere  varies  as  the  cube  of  its  diameter. 

17.  The  illumination  from  a  source  of  light  varies  inversely  as  the 
square  of  the  distance.  If  a  book,  now  10  inches  off,  be  moved  10(V5  — 1) 
inches  farther  away,  how  much  will  the  light  received  be  reduced  ? 

18.  Prove  that  if  x cc 0,  and  yccz,  then  x±yccz,  and  Vxy cc z, 

19.  Prove  that  if  x<xy,  and  zccu,  then  xz  qc  yu. 

20.  Prove  that  if  x  oc «/,  then  x»*  cc  y\ 


PROGRESSIONS  343 

XXV.    PROGRESSIONS 

ARITHMETIC  PROGRESSION 

517.  An  Arithmetic  Progression  is  a  series  (§  283)  in  which 
each  term,  after  the  iirst,  is  obtained  by  adding  to  the  preced- 
ing term  a  constant  number  called  the  Common  Difference. 

Thus,  1,  3,  5,  1,  9,  11,  •••  is  an  arithmetic  progression  in 
which  the  common  difference  is  2. 

Again,  12,  9,  6,  3,  0,  —  3,  •••  is  an  arithmetic  progression  in 
which  the  common  difference  is  —  3. 

An  Arithmetic  Progression  is  also  called  an  Arithmetic  Series. 

518.  Given  the  first  term,  a,  the  common  difference,  d,  and  the 
number  of  terms,  n,  to  find  the  last  term,  I. 

The  progression  is  a,  a  +  cZ,  a  +  2  d,  a  4-  3  d,  •  ••. 
We  observe  that  the  coeihcient  of  d  in  any  term  is  less  by  1 
than  the  number  of  the  term. 

Then,  in  the  nth  term  the  coefficient  of  d  will  be  n  —  1. 
That  is,  l  =  a-\-{n-l)d.  (I) 

519.  Given  the  first  term,  a,  the  last  term,  I,  and  the  number 
of  terms,  n,  to  find  the  sum  of  the  terms,  S. 

S  =  a-\-{a-\-d)-\-{a  +  2d)-\-'"+{l-d)-{-l. 
Writing  the  terms  in  reverse  order, 

S  =  l^{l-d)-\-(l-2d)^'"-\-{a-\-d)  +  a. 
Adding  these  equations  term  by  term, 

2^  =  (a  +  0  +  (a  +  0  +  («  +  0+  -  +(a  +  r)  +  (a  +  0. 
Therefore,     2S  =  n{a-^l),  and  ^  =  ^(a  + 1).  (II) 

The  first  term,  common  difference,  number  of  terms,  last  term,  and  sum 
of  the  terms,  are  called  the  elements  of  the  progression. 

Substituting  in  (II)  the  value  of  I  from  (I),  we  have 

>S  =  |[2a  +  (n-l)d].  >^ 


344  ADVANCED   COURSE  IN  ALGEBRA 

520.   Ex.     In  the  progression  8,  5,  2,  —1,  —4,  ••.,  to  27 
terms,  find  the  last  terra  and  the  sum. 
Here,  a  =  8,  d  =  5-8  =  -3,  n  =  27. 
Substitute  in    (I),   Z  =  8  + (27- 1)(-3)  =-70. 

Substitute  in  (II),  S=^~{^-10)=-  837.  ; 

Jj 

The  common .  difference  may  be  found  by  subtracting  the  first  term 
from  the  second,  or  any  term  from  the  next  following  term. 

EXERCISE  74 

In  each  of  the  following,  find  the  last  term  and  the  suxn : 

1.  5,  14,  23,  ...  to  18  terms.   ^  ^^  >  '^^  ^^"'fuiL 

2.  9,  2,   -  5,  ...  to  23  terms:  "^  ^  ,      '  ^  ^      ^ 
V  Z.    —51,   —43,   —35,  ...  to  15  terms. 

4.    --,-—,-  3,  ...  to  16  terms. 

4  8 

/  6.   ^   1    _  1    ...  to  28  terms. 
6   6        2 

6.  §,  il,  ^,  ...  to  25  terms. 
9'  36    18 

7.  -— ,  -—;  -1,  ...  to  39  terms.  y 

^^8.    -1,  -^,  -il,  ...to52terms.         ^'      .  ^l  t 

10'       6'       30'  v3)L^ 

9.   3a +  46,  8a +  2  6,  13  a,  ...  to  10  terms.      ^^V^  '   ^ 

v/     10.    ^^,  %  %  ...  to  9  terms.  '  "'•''' 
y^  3         6     3  J 

A^  521.  If  any  three  of  the  five  elements  of  an  arithmetic  pro- 
gression are  given,  the  other  two  may  be  found  by  substituting 
the  known  values  in  the  fundamental  formulae  (I)  and  (II),  and 
solving  the  resulting  equations. 

5  5 

1.   Given  a  =  —  -,  w  =  20,  8  =  —  \-\   find  d  and  I. 
o  o 

Substituting  the  given  values  in  (II), 

-^  =  10(^-^4-^),  or  -1  =  -^  +  ?;   then,  Z  =  5_l  =  |. 
3  ^3/'  6         3'  '        362 


PROGRESSIONS  846 


Substituting  the  values  of  a,  n,  and  I  in  (I), 

-  = \-19d',   whence,  19  d  =  — ,  and  d  =  -' 


2.    Given  d  =  -3,  l  =  -39,  >S  =  -264;   find  a  and  n. 
Substituting  in  (I), 

_39  =  a+(7i-l)(-3),  or  a  =  3n-42.  (1) 

Substituting  the  values  of  I,  S,  and  a  in  (II), 

-264  =  -(37i-42-39),  or  -  528  =  3^2-81  w, 
or      n2-27?i  =  -176. 


wv,                   27±V729-704      27  ±  5     -,^^1-, 
Whence,   n  =  ^-^ =  — ~-  — 16  or  11. 

Substituting  in  (1),  a  =  48 -42  or  33-42  =  6  or  -9. 
The  solution  is,  a  =  6,  ?i  =  16 ;    or,  a  =  —  9,  ti  =  11. 
The  significance  of  the  two  answers  is  as  follows  : 
If  a  =  6  and  n  =  16,  the  progression  is 

6,  3,  0,  -  3,  -  6,  -  9,   -  12,   -  15,   -  18,   -  21,   -  24,   -  27,   -  30, 
-33,   -36,   -39. 

If  a  =  —  9  and  72  =  11,  the  progression  is 

-  9,  -  12,   -  15,  -  18,  -  21,  -  24,   -  27,  -  30,  -  33,   -  36,   -  39. 

In  each  of  these  the  sum  is  —  264. 

113 
3.   Given  a  =  -,  d  =  — — ,  S  —  —  -\   find  I  and  n. 

Substituting  in  (I),  Z  =  |  +  («  - 1)  (-  X)  =  5^.  (1) 

Substituting  the  values  of  a,  S^  and  I  in  (II), 

3      n  /I  ,  5  —  n\  o         /9  —  ?i\  2     n         oa 

-2  =  2(3+^'  ""'  -^  =  '\-l2->  "'  »'-9"  =  36- 

Solving  this,  n  =  12  or  —  3. 

The  value  71  =  —  3  must  be  rejected,  for  the  number  of  terms 
in  a  progression  must  be  a  imsitive  integer. 

5-12  7 


Substituting  n  =  12  in  (1),  I  ■■ 


12  12 


346       ADVANCED  COURSE  IN  ALGEBRA 

Any  value  of  n  which  is  not  a  positive  integer  must  be  rejected,  together 
v^'ith  all  other  values  dependent  on  it. 

From  (I)  and  (II),  general  formulce  for  the  solution  of  exam- 
ples like  the  above  may  be  readily  derived. 

4.    Given  a,  d,  and  S ;  derive  the  formula  for  n. 

By  §519,  2jS  =  7il2a-\-(n-l)d'],  or  dji""  +  (;2a-d)n  =  2 S. 

This  is  a  quadratic  in  n;  solving  by  formula  (1),  §  450, 


_d-2a±  V(2 a ~  df  -^^dS 

ih  —  „ • 

2d 


EXERCISE  75 

^  ry^  ^  1-  Given  d  =  S,  1  =  147,  w  =^19 ;  find  a  and  S. 

2.  Given  d  =  -6,  w  =  14,  S  =  -  616  ;  find  a  and  L 

3.  Given  a  =-69,  n  =  16,  I  =  S6  ;  find  d  and  S. 

4.  Given  a  =  8,  w  =  25,  S  =  -  2500  ;  find  d  and  l. 

^  5.    Given  a  =-,  ?  =  -  — ,  ^  =  -  78  ffind  d  and  n. 
4  4 

6.  Given    I  =—,  w  =  25,  ^9  =  ?^ ;  find  a  and  d. 

4  4 

7.  Given  a  =  -  -,  d  =  _  A,  ^  ^  _  ^ ;  find  w  and  I. 

5'  10  5   ' 

8.  Given  a  =  -  -,  Z  =  — ,  (?  =  - ;  find  n  and  S. 

3  2  6 

9.  Given  d  = .  w  =  55,  S  =  -166;  find  a  and  I 

12 

10.  Given   ?  =  ^,  w  =  24,  ^  =  241  ;  find  a  and  d. 

11.  Given   ?  =  ^,  d  =  -,  8  =  —  -,  find  «  and  n. 

6  6  6 

12.  Given  a  =  --,  Z  =  -— ,  8  =  -^;  find  c?  and  w. 

5  10  2   ' 

13.  Given  a  =  -—,  n  =  21,  /S' =  —  :  find  cZ  and  l. 

22  22 

14.  Given   I  =  ^,  d  =  —,  8  =  -  —  ;  find  a  and  ?i. 

12  12  3 

16.    Given  «=-— ,  d  =  \  8  =  -  —  ;  find  ri  and  I. 

6  3  2  ' 


PROGRESSIONS  347 

16.  Given  a,  Z,  and  n  ;  derive  the  formula  for  d. 

17.  Given  a,  n,  and  S ;  derive  the  formul?e  for  d  and  I. 

18.  Given  d,  ?z,  and  /S' ;  derive  the  formulae  for  a  and  I. 

19.  Given  a,  d,  and   Z ;  derive  the  formulae  for  n  and  S. 

20.  Given  d,  Z,  and  n  ;  derive  the  formulae  for  a  and  /S^. 

21.  Given  Z,  n,  and  S ;  derive  the  formulae  for  a  and  d 

22.  Given  a,  (f,  and  /S ;  derive  the  formula  for  I. 

23.  Given  a,  Z,  and  8 ;  derive  the  formulae  for  d  and  n. 

24.  Given  cZ,  Z,  and  8 ;  derive  the  formulae  for  a  and  n. 

522.  Arithmetic  Means. 

To  find  an  arithmetic  progression  of  m  +  2  terms,  whose 
first  and  last  terms  are  two  given  numbers,  a  and  h,  is  called 
inserting  m  arithmetic  means  hetiwen  a  and  h. 

Ex.   Insert  5  arithmetic  means  between  3  and  —  5. 
We  find  an  arithmetic  progression  of  7  terms,  in  which  a  =  3, 
and  Z  =  —  5 ;  substituting  n  =  7,  a  =  3,  and  Z  =  —  5  in  (I), 

_5  =  3  +  6d,  or  cZ  =  --. 
'3 

The  progression  is  3,  -,  -,  -1,  --,   -~,  ~  5. 

523.  Let  X  denote  the  arithmetic  mean  between  a  and  h. 
Then,  x~a  =  b  —  x,  or  2x  =  a-^b. 

Whence,  ^  =  ^^- 

That  is,  tlie  arithmetic  mean  between  two  numbers  equals  one- 
half  their  sum. 

EXERCISE  76 


1.   Insert  7  arithmetic  means  between  4  and  10. 

5 
6 

7 
3 


Insert  6  arithmetic  means  between  —  -  and 

6  2 

7 

3.  Insert  9  arithmetic  means  between  —  and  6. 


33 
4.   Insert  8  arithmetic  means  between  —  3  and  —  —  • 


348  ADVANCED  COURSE  IN  ALGEBRA 


6.   Insert  5  arithmetic  means  between  -  and 

9  3 

3  29 

6.  How  many  arithmetic  means  are  inserted  between  —  -  and  — , 

9 
when  the  sum  of  the  second  and  last  is  -  ? 

5 

7.  If  m  arithmetic  means  are  inserted  between  a  and  6,  find  the  first 
three. 

Find  the  arithmetic  mean  between : 

8.  1|  and  -  2| .  9.    (3  m  +  n)2  and  (m  -  3  ny. 


x-l  x^-1 

524.   Problems. 

1.  The  sixth  term  of  an  arithmetic  progression  is  -,  and  the 

-I  p  D 

fifteenth  term  is  —     Find  the  first  term. 

The  common  difference  must  be  one-ninth  the  result  obtained  by 
subtracting  the  sixth  term  from  the  fifteenth. 

Then,  a  =  m-'-]  =  l- 

9V3      6/      2 

Again,  the  first  term  must  equal  the  sixth  term  minus  five  times  the 
common  difference. 

Then,  a  = =  —  -• 

'  6     2         3 

2.  Find  four  numbers  in  arithmetic  progression  such  that 
the  product  of  the  first  and  fourth  shall  be  45,  and  the  product 
of  the  second  and  third  77. 

Let  the  numbers  be  represented  by  x  —  3 ?/,  x  —  y,  x  +  y,  and  x  +  Sy, 
respectively. 

Then  by  the  conditions,    i 

^  1.^2  _     2/2  =  77. 

Solving  these  equations, 

x  =  9,  y  =  ±2;  or,  x=-9,  y  =  ±2  (§472). 

Then  the  nunfbers  are  3,  7,  11,  15  ;  or,  —  3,  —  7,  —  11,  —  15. 

In  problems  like  the  above,  it  is  convenient  to  represent  the  unknown 
numbers  by  symmetrical  expressions. 

Thus,  if  five  numbers  had  been  required  to  be  found,  we  should  have 
represented  them  hy  x  —  2  y,  x  —  y,  x,  x  -{■  y,  and  x'+  2}/. 


PROGRESSIONS  349 

EXERCISE  77 

13 

1.  The  seventh  term  of  an  arithmetic  progression  is ,  and  the 

7  ^ 

thirteenth  term  —  - .     Eind  the  twenty-second  term. 

2.  The  first  term  of  an  arithmetic  progression  is  1,  and  the  sum  of  the 
sixth  and  tenth  tferms  is  37.     Find  the  second,  third,  and  fourth  terms. 

Q 

3.  The  first  term  of  an  arithmetic  progression  of  eleven  terms  is  - ,  and 
the  seventh  term  —  3.    Find  the  sum  of  the  terms. 

4.  In  an  arithmetic  progression,  the  sum  of  the  first  and  last  terms  is 
two-ninths  the  sum  of  all  the  terms.     Find  the  number  of  terms. 

5.  How  many  positive  integers  of  three  digits  are  multiples  of  13  ? 
What  is  their  sum  ? 

6.  Find  five  numbers  in  arithmetic  progression  such  that  their  sum 
shall  be  25,  and  the  sum  of  their  squares  135. 

7.  Find  four  numbers  in  arithmetic  progression  such  that  the  product 
of  the  first  and  third  shall  be  —  21,  and  the  product  of  the  second  and 
fourth  24. 

8.  If  the  constant  difference  of  an  arithmetic  progression  equals  tvsrice 
the  first  term,  the  quotient  of  the  sum  of  the  terms  by  the  first  term  is  a 
perfect  square. 

9.  In  any  arithmetic  progression,  the  sum  of  the  first  m  terms,  less 
twice  the  sum  of  the  first  w  +  1  terms,  plus  the  sum  of  the  first  m  +  2 
terms,  equals  the  common  difference. 

10.  The  sum  of  the  first  ten  terms  of  an  arithmetic  progression  is  to 
the  sum  of  the  first  five  terms  as  13  to  4.  Find  the  ratio  of  the  fii*st  term 
to  the  common  difference. 

11.  The  sum  of  six  numbers  in  arithmetic  progression  is  36,  and  the 
sum  of  their  squares  is  286.     Find  the  numbers. 

'      12.  A  man  travels  -^—  miles.     He  travels  20  miles  the  first  day,  and 

A 

increases  his  speed  one-half  mile  in  each  succeeding  day.    How  many 
days  does  the  journey  require  ? 

13.  Find  three  numbers  in  arithmetic  progression,  such  that  the  square 
of  the  first  added  to  the  product  of  the  other  two  gives  16,  and  the  square 
of  the  second  added  to  the  product  of  the  other  two  gives  14. 

14.  A  traveller  sets  out  from  a  certain  place,  and  goes  3^  miles  the  first 
hour,  3f  the  second  hour,  4  the  third  hour,  and  so  on.  After  he  has  been 
gone  5  hours,  another  sets  out,  and  travels  8^  miles  an  hour.  After  how 
many  hours  are  the  travellers  together  ? 

(Interpret  the  two  answers. ) 


350  ADVANCED   COURSE  IN   ALGEBKA 

15.  Find  the  sum  of  the  terms  of  an  arithmetic  progression  of  11  terms, 
in  which  121  is  the  middle  terra. 

16.  A  man  climbing  a  mountain,  ascends  the  first  hour  1000  feet,  the 
second  hour  800  feet,  the  third  hour  600  feet,  and  so  on.  After  how 
many  hours  will  he  be  at  the  height  of  2800  feet  ? 

(Interpret  the  two  answers.) 

17.  If  a  person  saves  $  120  each  year,  and  puts  this  sum  at  simple 
interest  at  3|  per  cent  at  the  end  of  each  year,  to  how  much  will  his 
property  amount  at  the  end  of  18  years  ?  iLft-;44^ 

18.  There  are  12  equidistant  balls  in  a  straight  line.  A  person  starts 
from  a  position  in  line  with  the  balls,  and  beyond  them,  his  distance  from 
the  first  ball  being  the  same  as  the  distance  between  the  balls,  and  picks 
them  up  in  succession,  returning  with  each  to  his  original  position.  He 
finds  that  he  has  walked  6460  feet.     Find  the  distance  between  the  balls. 

19.  A  and  B  travel  around  the  world,  the  circuit  being  23661  miles. 
A  goes  east  one  mile  the  first  day,  two  miles  the  second  day,  three  miles 
the  third  day,  and  so  on.  B  goes  west  at  a  uniform  rate  of  20  miles  a  day. 
After  how  many  days  will  they  meet  ? 

(Interpret  the  negative  answer.) 

GEOMETRIC    PROGRESSION 

525.  A  Geometric  Progression  is  a  series  in  which  each  term, 
after  the  first,  is  obtained  by  multiplying  the  preceding  term 
by  a  constant  number  called  the  Ratio. 

Thus,  2,  6, 18,  54, 162,  •••  is  a  geometric  progression  in  which 
the  ratio  is  3. 

9,  3,  1,  -,  -,  •••  is  a  geometric  progression  in  which  the  ratio 

—  3,  6,  —12,  24,  —48,  •••  is  a  geometric  progression  in 
which  the  ratio  is  —  2. 

A  Geometric  Progression  is  also  called  a  Geometric  Series. 

526.  Given  the  first  term,  a,  the  ratio,  r,  and  the  number  of 
terms,  n,  to  find  the  last  term,  I. 

The  progression  is  a,  ar,  ar^,  ai^,  •••. 

We  observe  that  the  exponent  of  r  in  any  term  is  less  by  1 
than  the  number  of  the  term. 


PROGRESSIONS  351     ^ 

Then,  in  the  nth  term  the  exponent  of  r  will  be  n  —  1.        ^^    ~ 
That  is,  l=ar''-\  (I)    c^ 

527.  Given  the  first  term,  a,  the  last  term,  Z,  and  the  ratio,  r, 
to  find  the  sum  of  the  terms,  S.  4(9t         yd     '^ 

S  =  a  +X^  -\-ar^A h  c*?*"~^  +  ot^"~^  +  ^""^  (1) 

Multiplying  each  term  by  r, 

r/S  =  ^  +  ar^  +  ar'  H h  ar"-^  +  ow:^-i  4.  «?''*.  (2) 

Subtracting  (1)  from  (^,/aS  -  >S  =  ar^  -  a,  or  /S  =  ^El^. , 
But  by  (I),  §  526,         "^  ;'  r?  =  ar\  ^--Z^ 

Therefore,  ^^  ^  =  !lz:^.   v  (II) 

)  r-1      ■ 

The  first  term,  ratio,  number^  terms,  last  term,  and  sum  of  the  terms, 
.are  called  the  elements  of  the  progression. 

528.  Examples. 

1.  In  the  progression  3,  1,  -,  •••,  to  7  terms,  find  the  last 
term  and  the  sum. 

Here,  a  =  3,  r  =  -,  n  =1 ;  substituting  in  (I)  and  (II), 
o 

^3;      3^      243 

1     _1__3     _1._3         2186 
^_3      243        _729        _       729  _1093 

1_1       ~     _?     ~     _?     "243* 
3  3  3 

The  ratio  may  be  found  by  dividing  the  second  term  by  the  first,  or 
any  term  by  the  next  preceding  term. 

2.  In  the  progression  —2,  6,  —18,  •••,  to  8  terms,  find  the 
last  term  and  the  sum. 

Here,  a  =  —  2,  r  = =  —  3,  n  —  %;  therefore, 

—  z 

l  =  -2(-Sy  =  -2x(-  2187)  =  4374. 
^  ^  -  3  X  4374  -  (-  2)  ^  - 13122  +  2  ^  g^g^ 


352  ADVANCED   COURSE   IN   ALGEBRA 

EXERCISE   78 

In  each  of  the  following,  find  the  last  term  and  the  sum  of  the  terms 
—  2,  4,  ...  to  10  terms. 
3,  -  12,  -  48,  ...  to  6  terms. 


1.    1,  -  2,  4,  ...  to  10  terms.  «        3  .       4        .„  .  _,„ 

'         '    '  6.    —-,  1, —-,...  to  7  terms. 


/ 


3. 

_||,_^,...to9ter,„s. 

4. 

A,  ||,...  to  6  terms. 

5. 

-,  *,  2,  -..  to  7  terms. 

4'    '      3' 

16        4    1         ,    «, 

y  ~3'  3'  *"  ^^^*^^"^^- 

11  A,  ...to  5  terms. 

529.  If  any  three  of  the  five  elements  of  a  geometric  pro- 
gression are  given,  the  other  two  may  be  found  by  substituting 
the  given  values  in  the  fundamental  formulae  (I)  and  (II),  and 
solving  the  resulting  equations. 

But  in  certain  cases  the  operation  involves  the  solution  of 
an  equation  of  a  degree  higher  than  the  second ;  and  in  others 
the  unknown  number  appears  as  an  exponent,  the  solution  of 
which  form  of  equation  can  usually  only  be  affected  by  the  aid 
of  logarithms  (§  604). 

In  all  such  cases  in  the  present  chapter,  the  equations  may 
be  solved  by  inspection. 

1.   Given  a  =  —  2,  n  =  5,  I  =  —  32;  find  r  and  S. 
Substituting  the  given  values  in  (I),  we  have 

_32  =  -2r*;  whence,  r'  =  16,  or  r  =  ±2. 
Substituting  in  (II), 

If       r=     2,  /S  =  ^^~^P~/~^^=-64-f  2  =  -62. 

If       ,^_2,  ^^(-2)(-32)-(-2)^64  +  2^_ 
-2-1  -3 

The  solution  is,  r  =  2,  /S  =  -  62 ;  or,  r  =  - 2,  JS  =  -22. 

The  significance  of  the  two  answers  is  as  follows  : 

If  r=     2,  the  progression  is  —2,  —4,  —8,  —16,  —32,  whose  sum  is  —62. 

If  r=  —2,  the  progression  is  —2,     4,  —8,      16,  —32,  whose  sum  is  —  22. 


1640 

-1'- 

-3 

~    4 

729  ~ 

1 
3 

-1 

Z  +  9  = 

6560 

■  729' 

or, 

l  =  - 

1 

729 

PROGRESSIONS  868 


2.    Given  a  =  S    r  =  -i    S  =  ^^;  find  n  and  Z. 


Substituting  in  (II), 

Whence, 

Substituting  the  values  of  a,  r,  and  I  in  (I), 

Whence,  by  inspection,  w  —  1  =  7,  or  n  =  S. 

From  (I)  and  (II)  general  formulae  may  be  derived  for  the  solution  of 
cases  like  the  above. 

If  the  given  elements  are  n,  I,  and  8,  equations  for  a  and  r  may  be 
found,  but  there  are  no  definite  formulce  for  their  values. 

The  same  is  the  case  v^hen  the  given  elements  are  a,  n,  and  S. 

The  general  formulae  for  n  involve  logarithms ;  these  cases  are  dis- 
cussed in  §604. 

EXERCISE  79 

1.  Given  r  =  3,  w  =  8,  Z  =  2187  ;  find  a  and  S. 

2.  Given  a  =  6,  n  =  7,  Z  =  — ;  find  r  and  /S. 

243 

3.  Given  r  =  -5,  n  =  5,  S  =  -  1042  ;  find  a  and  I. 

4.  Given  a  =  - 3,  r  =  --,  I  =  -^:  find  n  and  S. 

'  2'         128' 

5.  Given  r  =  -2,  n  =  10,  S  =  -  ^^ ;  find  a  and  l. 

2     ' 

6.  Given  a  =  ^,  w  =  6,  Z  =  -  — ;  find  r  and  /S. 

2'  '  125' 

7.  Given  a  =-1  Z  =  -^,  S  =  -  —  ;  find  r  and  n. 

3'  64 '  192 ' 

8.  Given  a  =  -,r  =  -,S  =  ?^:  find  I  and  n. 

4'         4'  1024 ' 

9.  Given  I  =  384,  r  =  -  4,  S  =  ^^ ;  find  a  and  n. 

10.    Given  a  =  -,  Z  =  1458,  /S  =  ^^  ;  find  r  and  w. 
9  9 


:) 


354       ADVANCED  COURSE  IN  ALGEBRA 

11.  Given  a,  r,  and  S ;  derive  the  formula  for  I. 

12.  Given  a,  I,  and  S ;  derive  the  formula  for  r. 

13.  Given  r,  I,  and  S ;  derive  the  formula  for  a. 

14.  Given  r,  «,  and  I ;  derive  the  formulse  for  a  and  S. 

15.  Given  r,  n,  and  &' ;  derive  the  formulse  for  a  and  I. 

16.  Given  a,  n,  and  I ;  derive  the  formulse  for  r  and  S. 

630.   Sum  of  a  Geometric  Progression  to  Infinity. 

The  limit  (§  245)  to  which  the  sum  of  the  terms  of  a  decreas- 
ing geometric  progression  approaches,  when  the  number  of 
terms  is  indefinitely  increased,  is  called  the  sum  of  the  series 
to  infinity. 

Formula  (II),  §  527,  may  be  written 

-rl 


S 


1-r 


It  is  evident  that,  by  sufficiently  continuing  a  decreasing 
geometric  progression,  the  absolute  value  of  the  last  term  may 
be  made  less  than  any  assigned  number,  however  small. 

Hence,  when  the  number  of  terms  is  indefinitely  increased, 
I,  and  therefore  rl,  approaches  the  limit  0. 

/     Then,  the  fraction approaches  the  limit 

1  — r     ^^  1  — r 

/^  Therefore,  the  sum  of  a  decreasing  geometric  progression  to) 

(^infinity  is  given  by  the  formula  ^ 

1.   rind  the  sum  of  the  series  4,  — 
Here,  a  =  4,  r  =  — -• 
Substituting  in  (III),  S  = 


This  signifies  that,  the  greater  the  number  of  terms  taken,  the  more 
nearly  does  their  sum  approach  to  — ;  but  the  sum  will  never  exactly 
equal  this  value. 


1-r 

(III) 

ries4   -^    1*5   . 
"es  4,      -,  -, 

••  to  infinity. 

4        12 

'-1  ' 

PROGRESSIONS  355 

A  repeating  decimal  is  a  decreasing  geometric  progression, 
^  and  its  value  may  be  found  by  formula  (III). 

2.   Find  the  value  of  .85151  •••.  ■oooS'l^  ,o  /  -A/ 

We  have,  .85151  ...  =  .8  +  .051  +  .00051  +  ....       •  ^  ^^' 

The  terms  after  the  first  constitute  a  decreasing  geometric 

progression,  in  which  a  =  .051,  and  r  =  .01. 

Q  K  f  f  V       •     /TTTN    c        -051        .051      51       17 
Substituting  in  (III),  ^  =  ^_^  =  _  =  _  =  _. 

8       17  281 

Then,  the  value  of  the  given  decimal  is  tx  +  ^^j  or  — -•    . 

10      oo\)  oo\) 

^  EXERCISE  80 

\     Find  the  sum  of  the  following  to  infinity : 

\  1.  6,  -2,    1,    ....  5 


8     14     49 
5'   15'   90' 


^'  ^^'  ^'4'    •"•  ®-   "10'    25'    "125'    •••• 
Q         4        2          1  3     15      _J5^     _ 

^-    ~    '  "3'    "9'    **'•  4'   32'        256'       ' 

4   _25     25     _50  o    5      5       10 
•       6  '    9  '        27'    ""  ■   6'   27'   243' 

^       Find  the  values  of  the  following : 

9.  .8181  ....  11.  .91777  ....  13.  .23135135  .... 

10.  .629629  ....  12.  .75959  ....  14.  .587474  .... 

531.   Geometric  Means. 

To  find  a  geometric  progression  of  m  +  2  terms,  whose  first 
and  last  terms  are  two  given  numbers,  a  and  b,  is  called  insert- 
ing m  geometric  means  between  a  and  b. 

Ex.   Insert  5  geometric  means  between  2  and  — — . 
^  729 

We  find  a  geometric  progression  of  7  terms,  in  which  a  =  2, 

1 98  1 28 

and  I  =  —  ;  substituting  w  =  7,  a  =  2,  and  I  =  — -  in  (I), 

i  Zv  i ZJ 

II  =  2  r«  i  whence,  ,■«  =  ^,  and  r  =  ±  |. 


356  ADVANCED  COURSE  IN   ALGEBRA 

The  result  is  2,  ±^J,   ±1^,??,   ±^,^28^ 
'      3'  9'       27'  81'       243'  729 

532.   Let  X  denote  the  geometric  mean  between  a  and  6. 

Then,  ^  =  ^,  or  a;2  =  a6. 

a     X 

Whence,  x  =  Va6: 

That  is,  the  geometric  mean  between  tivo  numbers  is  equal  to 
the  square  root  of  their  product. 

EXERCISE  81 

o 

1.  Insert  8  geometric  means  between  -  and  —  192. 

2.  Insert  7  geometric  means  between  —  3  and  —  19683. 

3.  Insert  6  geometric  means  between  —  and  5120. 

16 

4.  Insert  4  geometric  means  between  —  -  and  — • 

6         729 

o 

5.  Insert  5  geometric  means  between  —  48  and  — — . 

256 

6.  Insert  3  geometric  means  between  —  and  — 

^  32  10 

7.  If  w  geometric  means  are  inserted  between  a  and  6,  what  are  the 
last  two  means  ? 

Find  the  geometric  mean  between  : 

9.   t±^^  and  ?^^. 


10.  a2  _  4  «5  +  4  52  and  4:0?  +  4:ah  +  h\ 

533.   Problem. 

Find  3  numbers  in  geometric  progression  such  that  their 
sum  shall  be  14,  and  the  sum  of  their  squares  84. 
Let  tlie  numbers  be  represented  by  a,  ar,  and  ar^. 


^,        ,                               r      a  +  ar -\- ar"^  =  14. 
Then,  by  the  conditions,  -{    „ 

U2  +  a2^2  +  Qj2^  ^  84. 

(1) 

(2) 

Divide  (2)  by  (1),                   a  -  ar  +  ar^  =  6. 

(3) 

Subtract  (3)  from  (1),                         2  ar  =  8,  or  r  =  -• 

(4) 

PROGRESSIONS  357 


Substituting  in  (1),  a  +  4  +  —  =  14,  or  a^  -  10  a  +  16  =  0. 
a 

Solving  this  equation,  a  =  8  or  2. 

Substituting  in  (4),  r  =  ^  or  -  =  -  or  2. 

8       2     2 

Then,  the  numbers  are  2,  4,  and  8. 


EXERCISE  82 

1.  What  number  must  be  added  to  each  of  the  numbers  a,  6,  and  c, 
so  that  the  resulting  numbers  shall  form  a  geometric  progression  ? 

4 

2.  The  sixth  term  of  a  geometric  progression  is  — ,  and  the  eleventh 

128  ' 

term Find  the  third  term. 

6561 

3.  Find  an  arithmetic  progression  whose  first  term  is  2,  and  whose 
first,  fourth,  and  tenth  terms  form  a  geometric  progression. 

4.  The  product  of  the  first  five  terms  of  a  geometric  progression  is  243. 
Find  the  third  term. 

5.  Find  four  numbers  in  geometric  progression  such  that  the  sum  of 
the  first  and  fourth  is  27,  and  of  the  second  and  third  18. 

6.  Find  six  numbers  in  geometric  progression  such  that  the  sum  of 
the  first,  third,  and  fifth  is  147,  and  of  the  second,  fourth,  and  sixth  294. 

7.  The  sum  of  the  terms  of  a  geometric  progression  whose  first  term 
is  1,  ratio  3,  and  number  of  terms  4,  equals  the  sum  of  the  terms  of  an 
arithmetic  progression  whose  first  term  is  4,  and  common  difference  4. 
Find  the  number  of  terms  in  the  arithmetic  progression. 

8.  A  man  who  saved  every  year  five-fourths  as  much  as  in  the  preced- 
ing year,  had  saved  in  four  years  $9225.  How  much  did  he  save  the  first 
year? 

9.  The  population  of  a  state  increases  from  100000  to  161051  in  five 
years.    What  is  the  rate  of  increase  per  year  ? 

10.  The  difference  between  two  numbers  is  72,  and  their  arithmetic 
mean  exceeds  their  geometric  mean  by  8.    Find  the  numbers. 

11.  The  sum  of  the  first  eight  terms  of  a  decreasing  geometric  progrea- 
sion  is  to  the  sum  to  infinity  as  16  to  25.     Find  the  ratio. 

12.  There  are  three  numbers  in  geometric  progression  whose  sum  is  Q\. 

Q  fi  7 

If  the  first  be  multiplied  by  - ,  the  second  by  - ,  and  the  third  by  - ,  the 

resulting  numbers  will  be  in  arithmetic  progression.     What  are  the 
numbers  ? 


^  HARMONIC   PROGRESSION 


358  ADVANCED   COURSE   IN  ALGEBRA 

13.  The  digits  of  a  number  of  three  figures  are  in  geometric  progres- 
sion, and  their  sum  is  14.  If  594  be  subtracted  from  the  number,  the 
digits  will  be  reversed.     Find  the  number. 

14.  The  mth  term  of  a  geometric  progression  is  p,  and  the  nth.  term  is 
q  ;  prove  that  the  first  term  is  ^'Vp^-^'q'"--^. 

15.  The  sum  of  three  rational  numbers  in  geometric  progression  is 

— ,  and  the  sum  of  their  reciprocals  —     Find  the  numbers. 
15  3 

16.  If  the  numbers  a,  b,  and  c  are  in  geometric  progression,  prove 

1  +  1  +  1  ^  «^  +  &^  +  c3 
a^     b^     c^  a%'^c^ 

17.  The  sum  of  the  first  four  terms  of  a  geometric  progression  is  45, 
and  of  the  first  lix  terms  189.     Find  the  first  term  and  the  ratio. 

18.  If  oj,  y,  and  z  are,  respectively,  the  pth,  gth,  and  rth  terms  of  a 
geometric  progression,  prove 

xQ-r  X  zP-^  =  yp-^. 

U       534.   A   Harmonic   Progression  is  a  series  of   terms  whose 
reciprocals  form  an  arithmetic  progression. 

Thus,  1,  -,  -,-,-,•••  is  a  harmonic  progression,  because  the 
3  5  7  9 
reciprocals  of  the  terms,  1,  3,  5,  7,  9,  •••,  form  an  arithmetic 
progression. 

A  Harmonic  Progression  is  also  called  a  Harmonic  Series. 

Any  problem  in  harmonic  progression,  which  is  susceptible 
of  solution,  may  be  solved  by  taking  the  reciprocals  of  the 
terms,  and  applying  the  formulae  of  the  arithmetic  progression. 

There  is,  however,  no  general  method  for  finding  the  sum  of 

the  terms  of  a  harmonic  series. 

2   2 
Ex.   In  the  progression  2,  -,  -,  •••  to  36  terms,  find  the  last 
,  3   5 

term. 

Taking  the  reciprocals  of  the  terms,  we  have  the  arithmetic 

.       13    5 

progression  -,  -,  -,  .... 

Here,  a  =  -,  d  =  1,  n  =  36. 


PROGRESSIONS  359 

Substituting  in  (I),  §  518,   Z  =  i+  (36  -  1)  x  1  =^- 

IThen,  —  is  the  last  term  of  the  given  harmonic  series. 
^'^  '  71  ^ 

535.  Harmonic  Means. 

To  find  a  harmonic  progression  of  m  +  2  terms,  whose  first 
and  last  terms  are  two  given  numbers,  a  and  6,  is  called  insert- 
ing m  harmonic  means  between  a  and  h. 

Ex.   Insert  5  harmonic  means  between  2  and  —  3. 

We  have  to  insert  5  arithmetic  means  between  -  and 

11  -   2  3 

Substituting  a  =  i  ?  =  -  ±,  «  =  7,  in  (1),  §  518, 
Z  o 

=  -  4-  6  a,  or  a  = 

3     2'  36 

Then  the  arithmetic  series  is 

1    13   2    1^    _1^    _]_    _1 
2'  36'  9'  12'       18'       36'       3* 

Therefore,  the  required  harmonic  series  is 

2,   ||,   I   12,    -18,    -^,    -3. 

536.  Let  X  denote  the  harmonic  mean  between  a  and  h. 

Then,  -  is  the  arithmetic  mean  between  -  and  — 
X  ah 

Then,  by  §  523,  1  =  5LJ  =  ^+^,  and  ^  =  1^- 
'    -^  'a;         2         2a6'  a  +  6 


EXERCISE  83 
In  each  of  the  following,  find  the  last  term : 

1.  ^    12     12        ^Q  22  terms. 
6'  43'   71 

2.  --,  --,  -—,  ...to  19  terms. 

8'       6'       11 


360  ADVANCED  COUKSE  m  ALGEBKA 


3.  -3,  2,  -,  ...  to  26  terms. 

14       2 

4.  — ,   — ,   — ,  •••  to  11  terms. 
11     39'    17 

5.  --,   --,   --,...  to  37  terms. 

6'       3'       9' 

6.  Insert  6  harmonic  means  between  2  and . 

9 

4  1 

7.  Insert  8  harmonic  means  between  —  and  — • 

5  5 

2  2 

8.  Insert  7  harmonic  means  between  -  and 

7  5 

Find  the  harmonic  mean  between  : 

9.  2  and  -3.  10.   ^  +  ^  and  «^^:l^. 
3              4  a-b  a^  +  b-^ 

11.  Find  the  (w  —  l)th  term  of  the  harmonic  progression  a,  b,  •••  to  n 
terms. 

12.  If  m  harmonic  means  are  inserted  between  a  and  6,  what  is  the 
third  mean  ? 

13.  The  first  term  of  a  harmonic  progression  is^j,  and  the  second  terra 
q.     Continue  the  series  to  three  more  terms. 

14.  The  arithmetic  mean  between  two  numbers  is  1,  and  the  harmonic 
mean  —  15.     Find  the  numbers. 

15.  The  fifth  term  of  a  harmonic  progression  is  — ,  and  the  eleventh 

term  —      Find  the  fifteenth  term. 
3 

16.  The    geometric    mean    between    two    numbers    is  4,    and    the 

1  fi  ^ 

harmonic  mean  —    Find  the  numbers. 
5 


17.  The  arithmetic  mean  between  two  numbers  exceeds  the  geometric 

mean  by  -,  and  the  geometric  mean  exceeds  the  harmonic  mean  by 

6  ^ 

-;-     Find  the  numbers. 

xo 

(Represent  the  sum  of  the  numbers  by  x,  and  their  product  by  y.) 

18.  Prove  that,  if  any  three  consecutive  terms  of  a  harmonic  progres- 
sion be  taken,  the  first  is  to  the  third  as  the  first  minus  the  second  is  to 
the  second  minus  the  third. 

19.  If  «2,  62^  and  c^  are  in  arithmetic  progression,  prove  that  &  +  c, 
c  -\-  a,  and  a  +  6  are  in  harmonic  progression. 


PROGRESSIONS  361 

20.  If  a,  6,  and  c  are  in  arithmetic  progression,  6,  c,  and  d  in  geomet- 
ric progression,  and  c,  d,  and  e  in  harmonic  progression,  prove  a,  c,  and  e 
in  geometric  progression. 

537.  Let  A,  6r,  and  H  denote  the  arithmetic,  geometric,  and 
harmonic  means,  respectively,  between  a  and  6. 
Then,  by  §§  523,  532,  and  536, 

«  +  &    ri_^/:T.  o^.i  7j-_  2a5  . 


z 
But,  ^x^^  =  a6=(V^)^. 


a  +  & 


Whence,  ^  X  11=  G\  or  G  =  V^  X  H. 

That  is,  ^/ie  geometric  mean  between  two  numbers  is  also  the 
geometric  mean  between  their  arithmetic  and  harmonic  means. 

538.   Let  a  and  b  be  two  positive  real  numbers. 
By  §  537,  their  geometric  mean  is  intermediate  in  value 
between  their  arithmetic  and  harmonic  means. 

But        c^  +  ?>      2ab  ^(a-{-by-4:ab^(a-b)\ 
'  2         a  +  b  2(a  +  b)  2(a4-&)' 

a  positive  number. 

Hence,  of  the  three  means,  the  arithmetic  is  the  greatest,  the 
geometric  next,  and  the  harmonic  the  least. 


862 


ADVANCED  COURSE   IN   ALGEBRA 


XXVI.    OONVERGENOY  AND  DIVERGENCY 
OF  SERIES 


539.   An  Infinite  Series  (§  283)  may  be  developed  by  Divi- 
sion, or  by  Evolution. 

Let  it  be  required,  for  example,  to  divide  1  by  1  —  a?. 

1-x 

X 

x  —  x^ 

The  quotient  is  obtained  in  the  form  of  the  infinite  series 

l  +  x+ic^H . 

Again,  let  it  be  required  to  find  the  square  root  of  1  +  ic. 


l^x 
1 


^2    -8^. 


2  + 


X  + 


^ 


2-i-x- 


x' 


The  result  is  obtained  in   the  form  of  the  infinite  series 

^2      8 

Infinite  series  may  also  be  developed  by  other  methods,  one 
of  the  most  important  of  which  will  be  considered  in  Chap. 
XXVII. 

540.   Convergent  and  Divergent  Series. 

An  infinite  series  is  said  to  be  Convergent  when  the  sum  of 
the  first  n  terms  approaches  a  fixed  finite  number  as  a  limit 
(§  245),  when  n  is  indefinitely  increased. 


CONVERGENCY  AND  DIVERGENCY  OF   SERIES    363 

This  limit  is  called  the  Value  of  the  Series. 
A  finite  series  may  be  regarded  as  a  convergent  series. 

An  infinite  series  is  said  to  be  Divergent  when  the  sum  of  the 
first  n  terms  can  be  made  numerically  greater  than  any  assigned 
number,  however  great,  by  taking  n  sufficiently  great. 

Consider,  for  example,  the  infinite  series 

developed  by  the  fraction (§  539). 

I.  Suppose  X  =  Xi,  where  x^  is  numerically  <  1. 

In  this  case,  the  given  series  is  a  decreasing  Geometric  Pro- 
gression ;  and  by  §  530,  the  sum  of  the  first  n  terms  approaches 

the  limit when  n  is  indefinitely  increased. 

J.    iCj 

That  is,  the  sum  of  the  first  n  terms  approaches  a  fixed  finite 
number  as  a  limit,  when  n  is  indefinitely  increased. 

Hence,  the  series  is  convergent  when  x  is  numerically  <  1. 

Let  us  take,  for  example,  x  =  .1. 

The  series  now  takes  the  form  1  +  .1  +  .01  +  ... ;  while  the  value  of 

the  fraction  from  which  the  series  was  developed  is  — = — ,  or  — . 

1  -  .1'        9 
In  this  case,  however  great  the  number  of  terms  taken,  their -sum  never 

exactly  equals  — ,  but  approaches  this  value  as  a  limit  (§  530). 
y 
Thus,  if  an  infinite  series  is  convergent,  the  value  of  the  series  (§  540) 
equals  the  value  of  the  expression  from  which  the  series  was  developed. 

II.  Suppose  X  =  fl^i,  where  x^  is  numerically  >  1. 

The  sum  of  the  first  n  terms  is  now  r"- 

1  +  a^  +  a^i^  +  -  +  xr"^  =  ^i^  (§  143),  tL^ 

By  taking  n  sufficiently  great,  -^ can  be  made  to  numer- 
al—1 

ically  exceed  any  assigned  number,  however  great. 

Hence,  the  series  is  divergent  when  x  is  numerically  >  1. 

Let  us  take,  for  example,  x  =  10. 

The  series  now  takes  the  form  1  +  10  +  100  +"•••;  while  the  value  of 

the  fraction  from  which  the  series  was  developed  is  ,  or  —  -• 

^  1-10'  9 


364       ADVANCED  COURSE  IN  ALGEBRA 

In  this  case,  the  greater  the  number  of  terms  taken,  the  more  does 
their  sum  diverge  from  the  value . 

Thus,  if  an  infinite  series  is  divergent,  the  greater  the  number  of  terms 
taken,  the  more  does  their  sum  diverge  from  the  value  of  the  expression 
from  which  the  series  was  developed. 

III.  Suppose  x  =  l. 

In  this  case,  each,  term  of  the  series  equals  1,  and  the  sum 
of  the  first  n  terms  equals  n;  and  this  sum  can  be  made  to 
exceed  any  assigned  number,  however  great,  by  taking  n  suffi- 
ciently great. 

Hence,  the  series  is  divergent  when  x  =  l, 

IV.  Suppose  x  =  —l. 

In  this  case  the  series  takes  the  form 

1-1+1-1  +  ...; 

and  the  sum  of  the  first  n  terms  is  either  1  or  0  according 
as  n  is  odd  or  even. 

If  the  sum  of  the  first  n  terms  of  an  infinite  series  neither 
approaches  a  fixed  finite  limit,  nor  exceeds  any  assigned  num- 
ber, however  great,  when  n  is  indefinitely  increased,  the  series 
is  called  an  Oscillating  Series. 

Hence,  the  series  is  an  oscillating  series  when  x==  —  l. 

541.  It  follows  from  §  540  that  an  infiiiite  series  cannot  he 
used  for  the  purposes  of  demonstration  unless  it  is  convergent. 

It  will  be  understood,  throughout  the  remainder  of  the  work,  that,  in 
every  expression  involving  a  convergent  infinite  series,  the  value  of  the 
series  is  meant.  For  example,  the  product  of  two  convergent  infinite 
series  will  be  understood  as  signifying  the  product  of  their  values. 

THEOREMS  ON  CONVERGENCY  AND  DIVERGENCY 

\  OF  SERIES  ,  ;   ;.  < -'   ' 

'^  542.  If  an  infinite  series  is  convergent,  the  last  term  approaches 
the  limit  zero,  when  the  number  of  terms  is  indefinitely  increased. 

Let  the  series  be  Wi  + 1^2  H h  ^^«  +  ^^«+i  H • 

By  §  540,  ?<i  + 1*2  H h  ^*n  and  u^  +  ^2  H h  ?*n  +  '^n+\  ap- 
proach the  same  finite  limit  when  n  is  indefinitely  increased. 


CONVERGENCY  AND  DIVERGENCY  OF  SERIES    365 

But  the  limit  of  the  difference  between  Wj  -f-  2^2  H h  ^«n  ^'^u[Z^ 

Ui  +  ^i2-\ h  Wn  4-  ^n+i  is  the  difference  of  their  limits  (§  255).   -h-^* 

Whence,  u^+i  approaches  the  limit  0  when  n  is  indefinitely 
increased. 


xj 


543.  An  infinite  series  is  convergent  if  the  sum  of  the  first  n 
terms  is  finite,  and  the  sum  of  any  finite  number  of  terms  com- 
mencing with  the  (n  +  l)th  approaches  the  limit  zero,  when  n  is 
indefinitely  increased. 

Let  C/ represent  the  sum  of  the  first  n  terms,  and  Fthe  sum 

of  themterms  w„_,.i-f  w.„+2H f-w„+^,  of  the  series  Wi  +  ?«2H • 

By  §  254,  the  limit  of  U-{-V,  when  n  is  indefinitely  increased, 

is  the  sum  of  the  limits  of  U  and  V. 

But  since  V  approaches  the  limit  0,  when  n  is  indefinitely 

increased,  U-\-  V  approaches  the  same  limit  as  U. 

Since  this  is  the  case  whatever  the  number  of  terms  in  V, 

U  must  approach  a  fixed  finite  limit  when  n  is  indefinitely 

increased;  and  the  series  is  convergent. 

Take,  for  example,  the  series  1  H 1 !-♦••. 

1-1 

2« 

The  sum  of  the  first  n  terms  is (§  143) ,  which  is  finite  however 

1  1 

great  n  may  he.  ^      2 

The  sum  of  the  terms  from  the  (w  +  l)th  to  the  (n  +  m)th  is 

2^  "^  2n+i  "^         ^2«+m-i'         2«\,       2"^  2"*-!/'       2"' 

which  approaches  the  limit  0  when  n  is  indefinitely  increased. 
Then,  the  series  is  convergent. 
(This  series  was  proved  convergent  in  §  540,  I.) 

544.  If  an  infinite  series  is  convergent  or  divergent,  it  will 
remain  convergent  or  divergent  after  any  finite  number  of  terms 
have  been  added  to,  or  subtracted  from  it. 

For  in  the  case  of  a  convergent  series,  the  sum  of  the  first  n 
terms  still  approaches  a  fixed  finite  limit  when  n  is  indefinitely 
increased. 


i-i 

2™ 

1-1 

2 


366       ADVANCED  COURSE  IN  ALGEBRA 

And  in  the  case  of  a  divergent,  the  sum  of  the  first  n  terms 
still  exceeds  any  assigned  number,  however  great,  when  n  is 
indefinitely  increased. 

Then,  in  testing  a  series  for  convergency  or  divergency,  we 
may  commence  at  any  assigned  term,  taking  no  account  of  the 
preceding  terms. 

^*  545.  If  all  the  terms  of  an  infinite  series  are  positive,  and  the 
sum  of  any  n  consecutive  terms  is  finite  however  great  n  may  be, 
the  series  is  convergent. 

For,  the  greater  the  number  of  terms  taken,  the  greater  will 
be  their  sum ;  but  this  sum  is  always  finite. 

Then,  when  n  is  indefinitely  increased,  the  sum  of  the  first  n 
terms  must  approach  sl  fixed  finite  limit;  and  the  series  is  con- 
vergent. 

Thus,  in  the  illustrative  example  of  §  543,  the  sum  of  the  first  n  terms 
\   is  finite,  however  great  n  may  be  ;  and  hence,  the  series  is  convergent. 

A  546.  The  following  theorem,  and  that  of  §  547,  are  of  great 
importance  in  testing  the  convergency  or  divergency  of  series : 
If,  commencing  with  a  certain  assigned  term,  each  term  of  a 
series  of  positive  terms  is  less  than  the  corresponding  term  of 
a  series  of  positive  terms,  ivhich  is  known  to  be  convergent,  the 
first  series  is  convergent. 

For  the  sum  of  the  first  n  terms  is  finite,  however  great  n  may 
be,  and  the  series  is  convergent  by  §  545. 

Ex.     Prove  the  infinite  series 

1  +  1+1  +  1  +  ... 

2     32     43  ^ 
convergent. 

Each  term,  commencing  with  the  third,  is  less  than  the  cor- 
responding term  of  the  series 

14_1-l14.14. 

which  was  proved  convergent  in  §  540,  I. 
Then,  the  given  series  is  convergent. 


N 


CONVERGENCY  AND   DIVERGENCY  OF   SERIES    367 

547.  If,  commencing  with  a  certain  assigned  term,  each  term  of 
a  series  of  positive  terms  is  greater  than  the  corresponding  term 
of  a  series  of  positive  terms,  ivhich  is  known  to  be  divergent,  the 
given  series  is  divergent. 

For,  the  sum  of  the  first  n  terms  may  be  made  to  exceed 
any  assigned  number,  however  great,  by  taking  n  sufficiently 
great. 

Thus,  commencing  with  tlie  second  term,  each  term  of  the  series 
1  +  2  +  22  +  23+... 
is  greater  than  the  corresponding  term  of  the  series 

1  +  1  +  1  +  1  +  -, 
which  was  proved  divergent  in  §  540,  III. 

Then,  the  given  series  is  divergent. 

(This  was  also  proved  in  §  540,  II.) 

,  548.    To  prove  the  infinite  series 

convergent  vjhen  k  is  >  1,  and  divergent  when  k  =  l  or  /c  <  1. 

I.  If  A;  is  >  1,  the  second  and  third  terms  are  together  < 
— -f— ,  or  — ;  the  next  four  terms  are  together  <— ;  the  next 

o 

eight  are  together  <  —  ;  and  so  on. 
Then,  the  series  is  less  than  the  series 

1|2,4,8,  i,l,l.i_i_ 


or 


l  +  2lri  +  (2^.J+(2^)+-; 


which  was  proved,  in  §  530,  to  approach  a  finite  limit  when 
the  number  of  terms  was  indefinitely  increased. 
Therefore,  the  given  series  is  convergent. 

II.   VLk  =  \,  the  series  becomes 

1  +  1  +  1  +  1  +  ....  (1) 

2     3     4  ^^ 


368  ADVANCED  COURSE  IN  ALGEBRA 

o  -t 

The  third  and  fourth  terms  are  together  >  -  or  - ;  the  next 
4       1  4       2 

four  terms  are  together  >  -  or  - ;  and  so  on. 

8       2 

Then,  by  taking  a  sufficiently  great  number  of  terms,  their 
sum  may  be  made  greater  than  any  assigned  number,  however 
great,  and  the  series  is  divergent. 

The  series  (1)  is  a  harmonic  series  (§  534). 

III.  If  A;  is  <  1,  each  term  of  the  given  series,  commencing 
with  the  second,  is  greater  than  the  corresponding  term  of  (1), 
and  the  series  is  divergent  (§  547). 

As  examples  of  the  above  general  theorem,  the  series 

22      32     42 
is  convergent ;  and  the  series 

a/2      VS      V4 
IS  divergent. 

549.  If,  commencing  with  a  certain  assigned  term,  each  term 
of  a  series  of  positive  terms  is  greater  than  some  assigned  finite 
number,  however  small,  the  series  is  divergent. 

For  the  sum  of  the  first  n  terms  can  be  made  to  exceed  any 
assigned  number,  however  great,  by  taking  n  sufficiently  great. 

Thus,  in  the  series  -  +  -  +  ?  +  7  +  ---, 

2      3      4      5 

each  term,  commencing  with  the  second,  is  greater  than  — 
Then,  the  series  is  divergent. 

550.  If,  in  two  series  of  positive  terms,  the  ratio  of  two  corre- 
sponding terms  is  ahcays  finite,  the  first  series  is  convergent  if 
the  second  is  convergent,  and  divergent  if  it  is  divergent. 

Let  the  series  be  Ui-\-U2-{-u^-{-  "-,  and  Vi4- ^2  +  ^3  + •••• 

I.    Let  the  second  series  be  convergent. 

Let  A:  be  a  finite  number  greater  than  the  greatest  of  the 

ratios  — ,  — ,  •••;  then,  —<k,  —<fc,  •.-,  —<k. 
Vi    v^       '  '  Vi        '   V2  v„ 


CONVERGENCY  AND   DIVERGENCY  OF  SERIES    369 

Then,     w^  <  kv-^,  u^  <  Jcv2,  •  •  •,  itn  <  ^^n- 
Adding  these  inequalities,  we  have 

(wi  +  U2  H h  u^)<'k(vi  -\-V2-\ h  v^). 

But  since  the  second  series  is  convergent,  Vi  + 1'2  H h  ^'n  is 

finite,  however  great  n  may  be ;  also,  k  is  finite. 

Then,  Wi  +  WgH h^n  is  finite,  however  great  n  may  be, 

and  the  first  series  is  convergent  (§  545). 

II.   Let  the  second  series  be  divergent. 

Let  Z:  be  a  finite  number  smaller  than  the  least  of  the  ratios 

^,  -^  ...;  then,  '^>k,  ^>A;,  ...,  -">fc. 

Then,  Ui  >  kv^,  ii^  >  ^^2}  -•-,  '^n>  ^'^n- 

Whence,  (wi  +  ^2  H h  w„)  >k(t\  +  V2-\ h  v^). 

Now,  since  the  second  series  is  divergent,  Vj  +  VgH- •••  +  v„ 
can  be  made  greater  than  any  assigned  number,  however  great, 
by  sufficiently  increasing  n. 

Then,  Wi  +  Wg  + hu^  can  be  made  greater  than  any  assigned 

number,  however  great,  by  sufficiently  increasing  n,  and  the 
first  series  is  divergent. 

1.   Prove  the  series 1 1 [- •••  convergent. 

1x2      2x3      3x4 

We  compare  the  series  with  the  series 

which  was  shown  to  be  convergent  in  §  548. 

The  ratio  of  the  nth.  term  of  this  to  the  nth.  term  of  the 

given  series  is  ^- — ,  or  ^^^  "^     ,  or  1  H — 

1  n^  n 

n(n  + 1) 

This  is  always  between  1  and  2,  and  is  therefore  finite  for 

every  value  of  n. 

Then  the  given  series  is  convergent. 


370  ADVANCED   COURSE   IN   ALGEBRA 

2.   Prove  the  series 1 1 h  •••  divergent. 

1x2      3x4     5x6 

We  compare  it  with  the  series 

1  +  1  +  1  +  -,  (1) 

which  was  shown  to  be  divergent  in  §  540,  III. 

The  ratio  of  the  wth  term  of  the  given  series  to  the  nth  term 
of  (1)  is  n'  ^^  1 

271(2.^-1/''''   2(^2-1^ 

This  is  always  between  -  and    -,  and  is  therefore  finite  for 

every  value  of  n ;  then  the  given  series  is  divergent. 

'^  551.  If,  commencing  with  a  certain  assigned  term,  the  ratio 
of  each  term  of  an  infinite  series  of  positive  terms  to  the  preceding 
term  is  numerically  less  than  a  fixed  positive  number,  which  is 
itself  less  than  unity,  the  series  is  convergent. 

Let  the  series  be  2^^  +  Wg  +  ^3  +  •*••  •       (1) 

Suppose  —Kk,  —<k,  •••,  -^^<k,  •••,  where  A;  is  <L 

Ui  U2  u,,_^ 

By  §  540,  I,  since  A;  is  <  1,  the  infinite  series 

1  +  A;  +  A;2  +  Ar^+... 
is  convergent. 

Then,  by  §  550,  since  u^  is  finite,  the  series 

^i(l  +  A:  +  A;2  +  A:«+-)  (2) 

is  convergent. 

Multiplying  together  the  first  w  —  1  of  the  above  inequalities 
(§  338),  we  have 

u^^h'"U^  <r-i;    whence,  '^<'k^-^,  or  ^^<u,lz^-\ 


rj. 


U-^U-2  ' ' '  U^_-y 

That  is,  the  nth  term  of  (1)  is  less  than  the  nth  term  of  (2) ; 
and  by  §  546,  series  (1)  is  convergent. 

The  ratio  of  Un+i  to  Un  is  called  the  Eatio  of  Convergency  of  the  infinite 
series  mi  +  W2  +  W3  +  •••. 

02      03      2* 
Ex.   Prove  the  series  1  +  .-^  +  ,-5  +  iT  H convergent. 

|2      [3      [4 


CONVERGENCY   AND   DIVERGENCY   OF   SERIES    371 
The  ratio  of  the  (n  +  l)th  term  to  the  rith.  is 

2n+l 

\n-hl  2 
or 


2~     '  71  +  1 

\^ 

2 
If  71  =  2,  the  ratio  equals  -;  and  for  any  vahie  of  n>2,  the 

2  ^ 

ratio  is  <  -• 
o 

Then,  commencing  with  the  fourth  term,  the  ratio  of  each 
2 
term  to  the  preceding  is  <  -,  and  the  series  is  convergent. 

-^  552.  If,  commencing  with  a  certain  assigned  term,  the  ratio 
of  each  term  of  an  infinite  series  of  positive  terms  to  the  preced- 
ing term  is  either  equal  to,  or  greater  than,  unity,  the  series  is 
divergent. 

For,  commencing  with  the  assigned  term,  each  term  of  the 
series  is  either  equal  to,  or  greater  than,  the  assigned  termj 
and  the  series  is  divergent  by  §  549. 

2  2^  2^ 

Ex.   Prove  the  series + 1 1 divergent. 

1x2     2x3     3x4 

The  ratio  of  the  (71  +  l)th  term  to  the  Tith  is 

2n+l 
(n  +  l)(7.  +  2)       ^^       271 


2"  71  -f-  2 


n  (n  + 1) 

If  n  =  2,  the  ratio  equals  1 ;  and  for  any  value  of  n>2,  the 
ratio  is  >  1. 

Then,  commencing  with  the  third  terra,  the  ratio  of  each 
term  to  the  preceding  is  equal  to,  or  greater  than,  1,  and  the 
series  is  divergent. 

553.  The  method  of  §  551  does  not  apply  when  the  ratio  of 
the  (n  +  l)th  term  to  the  nth  is  less  than  1,  but  approaches 
the  limit  1  when  n  is  indefinitely  increased. 


372       ADVANCED  COURSE  IN   ALGEBRA 

For  in  this  case  the  ratio  will  not  always  be  less  than  a  fixed 
positive  number,  which  is  itself  less  than  1. 

In  such  cases,  the  convergency  or  divergency  of  the  series 
must  be  determined  by  other  tests. 

Consider,  for  example,  the  series 

1  +  1  +  1+1+.... 

The  ratio  of  the  (n  +  l)th  term  to  the  nth  is 

-J—  2-1 

2n  +  l     2n  — 1  n 


2»+l     2+1 
n 


This  is  always  <  1,  but  approaches  the  limit  1  when  n  is 
indefinitely  increased. 

The  series  is  divergent,  which  may  be  shown  as  in  §  548,  II. 

Q    554.   If  an  infinite  series  of  positive  terms  is  convergent,  it  will 
,  remain  so  after  the  signs  of  any  of  its  terms  have  h^en  changed. 
For  the  sum  of  the  first  n  terms  will  still  be  finite,  however 
great  n  may  be,  and  the  series  is  convergent  by  §  545. 

It  follows  from  the  above  that  the  theorems  of  §§  546 
and  551  hold  when  any  or  all  of  the  terms  are  negative ;  and 
also  the  first  statement  of  §  550  when  all  the  terms  are 
negative. 

555.   It  follows  from  §§  551,  552,  and  554  that 
I.    If  the  ratio  of  the  (n  +  V)th  term  of  an  infinite  series  to  the 
nth  term  approaches  a  limit  numerically  <  1,  when  n  is  indefi- 
nitely increased,  the  series  is  convergent. 

2 
Thus,  in  the  example  of  §  551,  • approaches  the  limit  0,  when  n  is 

indefinitely  increased. 


<  Then,  the  series  is  convergent. 

II.    If  the  ratio  of  the  (n  -f  l)th  term  of  an  infinite  series  to 
y       the  nth  term  approaches  a   limit  numerically   >1,  ivhen  n  is 
indefinitely  increased,  the  series  is  divergent. 


/ 


CONVERGENCY  AND  DIVERGENCY  OF  SERIES    373 

Thus,  in  the  example  of  §  552,  the  ratio  of  the  (n  +  l)th  term  to  the 

nth  term  can  be  written ;  and  this  approaches  the  limit  2  when  n  is 

indefinitely  increased.      ^  '^  ^ 

Then,  the  series  is  divergent. 

If  the  ratio  of  the  (w  +  l)th  term  to  the  nth  term  approaches  the  limit 
1,  when  n  is  indefinitely  increased,  other  tests  must  be  applied  to  deter- 
mine whether  the  series  is  convergent  or  divergent  (§  553). 

556.  If  the  terms  of  an  infinite  series  are  alternately  positive 
and  negative,  and  each  term  numerically  less  than  the  preceding 
term,  the  series  is  convergent  if  the  nth  term  approaches  the  limit 
0  when  n  is  indefinitely  increased. 

Let  the  series  be  ii-^^  —  U2  -[-  u^  —  Uj^+  •  •  •. 
It  may  be  written  in  the  forms 

{u^  -  U2)  -f-  (us  -  W4)  +  (%  -  -Me)  +  •",  (1) 

and  u-L  —  {u2  —  u^  —  (u^  —  u^) •  (2) 

Since  each,  of  the  expressions  Ui  —  u^,  u^  —  u^,  etc.,  is  posi- 
tive, it  is  evident  from  (1)  and  (2)  that  the  sum  of  the  first  n 
terms  is  positive,  and  <  u^. 

That  is,  the  sum  of  the  first  n  terms  of  the  given  series  is 
finite,  however  great  n  may  be. 

The  sum  of  the  m  terms  commencing  with  the  {n  +  l)th  is 

the  upper  or  lower  sign  being  taken  before  u^+m  according  as 
m  is  odd  or  even. 

The  expression  in  parentheses  can  be  written  in  the  forms 

and  U^+i  -  K+2  -  Un+s)  -  (^n+i  —  Wn+s)  +  ' ' '  ; 

which  shows  that  it  is  positive  and  <  w„+i. 

Now,  if  n  is  indefinitely  increased,  ?<„+i,  and  therefore 
expression  (3),  approaches  the  limit  0. 

Thus,  the  sum  of  any  finite  number  of  terms  commencing 
with  the  (7i4-l)th  approaches  the  limit  0  when  n  is  indefinitely 
increased,  and  the  series  is  convergent  by  §  543. 


374  ADVANCED   COURSE   IN   ALGEBRA 

As  an  example  of  the  above  theorem,  the  series 

1-1  +  ^-1+... 

2     3     4^ 
is  convergent. 

"*i     557.    To  prove  the  infinite  series 

convergent  when  x  is  numerically  <  1. 

The  ratio  of  the  (n  +  l)th  term  to  the  nth.  term  is 


71      71  —  1  n  \        ^, 

which  approaches  the  limit  x  when  n  is  indefinitely  increased. 
Then,   if  x  is   numerically   <  1,  the   series   is   convergent 
(§  555,  I). 

558.    To  prove  the  infinite  seyies 

convergent  for  every  value  of  x.  a?" 

The  ratio  of  the  (n  +  l)th  term  to  the  nth.  is 


\n 


\n-l 
which  approaches  the  limit  0  when  ^i  is  indefinitely  increased. 
Then,  the  series  is  convergent  (§  555,  I). 

\       559.    To  prove  the  infinite  series 

[2  [3 

lohere  n  is  any  positive  fraction,  or  negative  integer  or  fraction, 
convergent  when  x  is  nuTnerically  <  1. 

By  §  287,  the  ratio  of  the  (r  +  l)th  term  to  the  rth  term  is 

n(n-l)^"{n-r-\-V)  ^,  _^  y?(n  -  1)  ■■■(n  ~r +  2)  ^.^^ 

\r  '  \r  —  l 


CONVERGENCY  AND  DIVERGENCY  OF  SERIES    375 


n  —  r-\-l^         fn-\-l 


That  is,  ■ — X,  or 


-1^. 


r 

If  n  does  not  equal  —  1,  this  approaches  the  limit  —  x  when  ' 
r  is  indefinitely  increased ;  and  the  series  is  convergent  if  x  is 
numerically  <  1  (§  555,  I)., 

If  n  =  —  1,  the  series  takes  the  form  1  —  x-{-x' —  x^ -\ ,    . 

which  is  convergent  when  x  is  numerically  <  1  by  §  556. 

560.  Tlie  infinite  series 

a  +  bx-\-cx^-{-dx^-] ^kx^-^ +  laf -\- •" 

is  convergent  when  the  numerical  value  of  x  is  taken  sufficiently 

small,  and  for  ariy  numerically  smaller  value  of  x,  including  zero. 

Ix^  Ix 

For  the  ratio  of  the  (n  +  l)th  term  to  the  nth  is  z— ^,  or  — • 

Whatever  the  values  of  k  and  I,  x  may  be  taken  so  small  that 

—  shall  always  be  numerically  <  g,  where  q  is  numerically  <  1. 
k 

Then,  for  this  value  of  x  the  series  is  convergent. 

And  by  §  546,  it  is  convergent  for  any  numerically  smaller 
value  of  X,  including  0. 

561.  If  X  has  any  value  (not  including  zero)  which  makes 
the  series  a  +  bx  +  cx' -{- ...  (1) 

convergent,  a  +  &a;  +  caj^  +  ...,  and   therefore  bx  +  cx^ -\ ,  or 

x(b-\-cx-^  '"),  is  finite,  for  this  value  of  x. 

Then,  since  x  is  not  zero,  6  +  ccc  +  •••  is  finite  for  this  value 

of  X ;  and  x(b-{-cx-\ ),  or  5ic  +  ca^  H approaches  the  limit  0 

when  x  is  indefinitely  decreased. 

Since  6  -f  ca;  -f-  •  •  •  is  finite  for  any  value  of  a;,  except  0,  which 
makes  series  (1)  convergent,  it  is  itself  convergent  for  this 
value  of  X  (§  545). 

562.  Absolutely  and  Conditionally  Convergent  Series. 

A  convergent  infinite  series,  having  negative  terms,  is  said  to 
be  absolutely  convergent  when  it  remains  convergent  after  the 
signs  of  the  negative  terms  are  changed. 


376  ADVANCED  COURSE  IN  ALGEBRA 

It  is  said  to  be  conditionally  convergent  when  this  is  not  the 
case. 

Thus,  the  convergent  infinite  series  (§  556) 

if  the  signs  of  the  negative  terms  are  changed,  becomes 

1  +  1.1-1.... 

which  was  proved  divergent  in  §  548. 

Then,  the  series  (1)  is  conditionally  convergent. 

In  a  conditionally  convergent  series,  the  sum  of  the  terms 
approaches  a  different  limit  by  a  different  arrangement  of  the 
terms. 

Thus,  in  series  (1),  we  may  write  the  terms 


,0.5. 


('-M)-(i-r(i-l)--- 

which  shows  that  the  sum  of  the  terms  is  <(1~q  +  q) 
l^Q  may  also  write  the  terms 

M-i)+G-l-i)H^n-j)-- 

which  shows  that  the  sum  of  the  terms  is  >  -  • 

y        Then  the  terms  of  a  series  cannot  be  arranged  in  any  desired 
\      order  unless  the  series  is  absolutely  convergent. 


EXERCISE  84 
Expand  each  of  the  following  to  four  terms : 

*l  +  2ac                           '3-6x2  +  7x3 
1  —  5  x2 

7. 
8. 
9. 

\/9  a*  -  3  62. 

2.    ■  .             .  --  . 

l-5x-2x2 

3            ^^ 

6.    Vl  +  3x. 

\/8x8-l. 

6.    Vx2  +  x2/  +  y2. 

y/a^  +  2  &. 

2  +  6  X  -  x2 

CONVERGENCY   AND   DIVERGENCY   OF   SERIES    377 

Determine  whether  the  following  infinite  series  are  convergent  or 
divergent : 

10,    1  +  1  +  1  +  1  +  ....  (oU  17     1  + A +  _!!_+... 

13.    l  +  2!  +  33  l+2_3     3^     43 

22^32     42^  2.3     3.4     4.5 

15.  1  +  L^  +  L^5        ._  22     l__l„4._i !_+.... 

33. 63. 6- 9  3     2.32     3.33     4.34^ 

16.  1  +  1  +  1+....  23.  ?  +  -§_  + -^  + -A. +  .... 
3^6^9^  1^1.2     2.3      3.4^ 

Determine  for  what  values  of  x  each  of  the  following  infinite  series  is 
convergent  or  divergent : 

24.    1  +  22X  + 32^:2  +  ....  25.    1+-^  +  -^  +  -^  +  .... 

26.   Prove  the  infinite  series 

1    +^JL_  +  ,i    +... 


1  +  X     1  +  a:2     1  +  x3 

convergent  when  a:  is  >  1,  and  divergent  when  x  —  \  or  aj  <  1. 

27.   Prove  the  infinite  series 

_1_  +  -JL-  +  __^  +  ... 
1  +  x     l  +  a;2^1+x4 

convergent  for  every  value  of  x. 

In  determining  the  convergency  or  divergency  of  a  series,  it  is  usually 
best  to  commence  with  the  tests  of  §  655  ;  if  the  limit  approached  is  1,  then 
other  methods  may  be  tried. 


378  ADVANCED   COURSE  IN  ALGEBRA 

XXVII.    UNDETERMINED   COEFFICIENTS 

THE  THEOREM  OF  UNDETERMINED  COEFFICIENTS 

563.   An  important  method  for  expanding  expressions  into 

series  is  based  on  the  following  theorem : 

Let  the  members  of  the  equation 

A  +  Bx+  C:»?  +  Dx^  +  '"  =  A  +  B'x  +  Ox"  +  Dy? ^-  '"     (1) 

be  injinite  series  which  are  equal  for  any  value  of  x  which  makes 
them  both  convergent  ;  then^  the  coefficients  of  like  powers  ofx  in 
the  series  will  be  equal;  that  is, 

A  =  A',  B  =  B',  C=a,etc. 

By  §  560,  each  series  is  convergent  for  a  certain  finite  value 
of  X,  and  for  all  smaller  values ;  and  since  this  is  the  case,  each 
of  the  expressions 

Bx-j-Gx"-^-"  3indB'x+C'x'+'" 

approaches  the  limrt  0  when  x  is  indefinitely  decreased  (§  561). 

Then,  the  given  series  approach  the  limits  A  and  A',  respec- 
tively, when  X  is  indefinitely  decreased. 

Now,  since  the  given  series  are  functions  of  x,  which  are 
equal  for  every  value  of  x  which  makes  them  both  convergent, 
their  limits  when  x  is  indefinitely  decreased  are  equal  (§  252). 

Therefore,  A  =  A\ 

and  hence  Bx-{-Cx^-\ =  B'x  -f  Ox^  -\ . 

Since  x  is  not  0,  we  may  divide  through  by  x. 

Then,  B+Gx+"'  =  B^  +  ax  +  ":  (2) 

By  §  561,  each  member  of  (2)  is  convergent  for  the  same 
values  of  x  as  the  corresponding  member  of  (1). 

That  is,  (2)  is  satisfied  by  any  value  of  x  which  makes  both 
members  convergent;  and  letting  x  be  indefinitely  decreased, 
we  have  j^  ^  ^r 


UNDETERMINED   COEFFICIENTS 


379 


Proceeding  in  this  way,  we  may  prove  C  =  C\  etc. 

The  above  proof  holds  when  either  or  both  of  the  given  series  are  finite. 


564.    1.   Expand 


EXPANSION  OF  FRACTIONS 


in  ascending  powers  of  x. 


I_2a;  +  3x2 

We  saw,  in  §  539,  that  a  fraction  could  be  expanded  into  a 
series  by  dividing  the  numerator  by  the  denominator. 

We  therefore  know  that  the  proposed  expansion  is  possible. 
Assume  then 

2  _  3  a;2  -  a;3 


=  A+Bx+Cx'-{-D3?  +  Ex''  + 


(1) 


l-2i«  +  3a;2 

where  A,  B,  C,  D,  E,  -",  are  numbers  independent  of  x. 

Clearing  of  fractions,  and  collecting  the  terms  in  the  second 
member  involving  like  powers  of  x,  we  have 


2-3x'-a^  =  A-{-    B 


x^    G 

x'-{-    D 

a^+    E 

-2B 

-2C 

-2D 

+  3^ 

-{-SB 

+  30 

^+ 


(2) 


A  vertical  line,  called  a  bar,  is  often  used  in  place  of  parentheses. 
Thus,  +     ^  I  a;  is  equivalent  to  (^  —  2  A)  x. 

-2a\ 

Equations  (1)  and  (2)  are  satisfied  when  x  has  any  value 
which  makes  the  second  member  a  convergent  series. 

Then,  by  §  563,  the  coefficients  of  like  powers  of  x  in  (2) 
must  be  equal ;  that  is, 
A=     2. 
B-2A=     0;  or,  B  =  2A  =4. 

(7_2B  +  3^  =  -3;  or,  0  =  2  5-3  ^-3  =  -1. 
i)_2  0+3  5  =  -l;  or,  2)  =  2  0-3J5-l=-15. 
^_2D  +  30=     0;  OT,E=2D-3C        =-27;  etc. 

Substituting  these  values  in  (1),  we  have 

2-Sx'-x^      2^4.x-a^-15a^-27x'--., 
The  result  may  be  verified  by  division. 


380 


ADVANCED  COURSE  IN  ALGEBRA 


The  series  expresses  the  value  of  the  fraction  only  for  such  values  of  x 
as  make  it  convergent  (§  540). 

If  the  numerator  and  denominator  contain  only  even  powers 
of  x^  the  operation  may  be  abridged  by  assuming  a  series  con- 
taining only  the  even  powers  of  x. 


Thus,  if  the  fraction  were 


2  +  4a^-a;^ 


we   should  assume 


it  equal  to  ^  +  B:^  +  Cx''  +  Dx^  +  Ex^  + .... 

In  like  manner,  if  the  numerator  contains  only  odd  powers 
of  £17,  and  the  denominator  only  even  powers,  we  should  assume 
a  series  containing  only  the  odd  powers  of  x. 

If  every  term  of  the  numerator  contains  ic,  we  may  assume  a 
series  commencing  with  the  lowest  power  of  x  in  the  numerator. 

If  every  term  of  the  denominator  contains  a;,  we  determine 
by  actual  division  what  power  of  x  will  occur  in  the  first  term 
of  the  expansion,  and  then  assume  the  fraction  equal  to  a  series 
commencing  with  this  power  of  a?,  the  exponents  of  x  in  the 
succeeding  terms  increasing  by  unity  as  before. 

1 


2.   Expand 


3a^-a^ 


in  ascending  powers  of  x. 


Dividing  1  by  3  a^,  the  quotient  is 
"We  then  assume, 


Z^-^     ^^^ 

-(—   JL_/a/ 

^^  ^  n 

—  j^ 

•^  T-  ^-^ 

Clearing  of  fractions. 

1=3^+35 

0^  +  3(7 

x'  +  SD 

o^  +  3E 

-    A 

-    B 

-    G 

-    D 

Equating  coefficients  of  like  powers  of  a;. 

3^  =  1 

35-J.  =  0 

3(7-5  =  0, 

3i)-C  =  0 

' 

3E-1 

D  =  0 

et 

c. 

(3) 


x*  + 


UNDETERMINED   COEFFICIENTS  '     881 


mence,  ^  =  |,  B  =  |,  C^l^,  i>  =  i,  E^^^,etc. 


Substituting  in  (3),  we  have 


JL  «c        I    "^        III      ^^      I 


3aj2-aj3      3    '    9    '  27  '  81  '  243 

In  Ex.  1,  E  =  2D  —  3C]  that  is,  the  coefficient  of  x*  equals 
twice  the  coefficient  of  the  preceding  term,  minus  three  times 
the  coefficient  of  the  next  but  one  preceding. 

It  is  evident  that  this  law  holds  for  the  succeeding  terms  ; 
thus,  the  coefficient  of  a;^  is  2  x  (-  27)  -  3  x  (- 15),  or  -  9. 

After  the  law  of  coefficients  has  been  found  in  any  expansion, 
the  terms  may  be  found  more  easily  than  by  long  division ; 
and  for  this  reason  the  method  of  §  564  is  to  be  preferred  when 
a  large  number  of  terms  is  required. 

The  law  for  Ex.  2  is  that  each  coefficient  is  one-third  the 
preceding. 

EXERCISE  85 


Expand  each  of  the  following  to  five  terms  : 

l+4:x-x^  9      x^-Sx 

l-x  +  3x2*            *   3-x-2x3  ""   2x^  +  x* 

-x  +  Sx^          jQ        2  -  3  x2  j^    34-5X- 

1+2x2    *             ■    3-2x  +  x3'  '    x2-3x3  +  x* 

1  +  4x2              ^^      2  +  X  -  3  x8  jg     l-4xg  +  6x8 

1  +  4  X  -  x2*               ■    1  -  4  X  +  5  x2'  *     X  +  2  x2  -  x8  * 

a;  -  7  x^  -  4  x^      ^2        3-4xg  ^g      l-4x  +  2x« 

6  x2       "■    1  _  5  X  -  2  x2  *         '    2  +  x2  -  3  x3*  *   2  x^  -  3  x»  -  x^* 

EXPANSION  OF  SURDS 


1-x 

l-6x 

l  +  4x 

4  +  x2 

1  -  3  x2 
2x 

565.  Ex.     Expand  Vl  —  a;  in  ascending  powers  of  x. 
We  saw,  in  §  539,  that  the  square  root  of  an  imperfect  square 
could  be  expanded  into  a  series  by  Evolution. 

We  therefore  know  that  the  proposed  expansion  is  possible. 
Assume  then, 

VT^^  =  A  +  BX+  Cx"  -\-  Dx^  -i-  Ex^  -{-  ....  (1) 


382 


ADVANCED  COURSE  IN  ALGEBRA 


Squaring  both  members,  we  have  by  §  134, 


l-x  =  A' 


+  2AB 


x-\-      B^ 

+  2AC 


x" 


+  2AD 
+  2BC 

Equating  coefficients  of  like  powers  of  x, 
A'=      1;  or,  ^  =  1. 
2AB  =  -1;  or,  B  =  - 

B'-{-'2AC=:     0;  or,  C=- 

2AD-^2BG=     0;or,  X>  =  - 

C'  +  2AE-\-2BD=     0;  or,  ^  =  - 

2  A 

Substituting  these  values  in  (1),  we  have 

2      8      16     128 
The  result  may  be  verified  by  Evolution. 


xl+  0' 
-{-2AE 
+  2BD 


»*  + 


1 

2A 

1 

2* 

B' 
2  A 

1 

8' 

BG 
A 

1 
16* 

G'  +  2BD 

128' 


etc. 


The  series  expresses  the  value  of  Vl  —  a;  only  for  such  values  of  x  as 
make  it  convergent. 

EXERCISE  86 


Expand  each  of  the  following  to  five  terms : 


(2 


vi^vV 


1.    VI  -  4  a;. 


2.    V1  +  5X. 


3.  Vl  +  2  X  -  QoK 

4.  Vl  -  a;  +  3  ic2. 


5.    VI +  3  a;. 


6.    Vl  -  X  -  2  x2. 


PARTIAL  FRACTIONS 

566.  If  the  denominator  of  a  fraction  can  be  resolved  into 
factors,  each  of  the  first  degree  in  x,  and  the  numerator  is  of  a 
lower  degree  than  the  denominator,  the  Theorem  of  Undeter- 
mined Coefficients  enables  us  to  express  the  given  fraction  as 
the  sum  of  two  or  more  imrtial  fractions,  whose  denominators 
are  factors  of  the  given  denominator,  and  whose  numerators 
are  independent  of  x. 


UNDETERMINED   COEFFICIENTS  383 

567.   Case  I.     No  factors  of  the  denominator  equal. 

1.   Separate  '  ^"*" into  partial  fractions. 

^  (3a;-l)(5a;  +  2) 

Assume -—- ^  =  z r  +  p — r^'  (■•■) 

where  A  and  B  are  numbers  independent  of  x. 
Clearing  of  fractions,  we  have 

l^x-{-l  =  A(hx  +  2)  +  B(Zx-l) 

=  {^A  +  ^B)x^2A-B.  (2) 

The  second  member  of  (1)  must  express  the  value  of  the 
given  fraction  for  every  value  of  x. 

Hence,  equation  (2)  is  satisfied  by  every  value  of  x]  and  by 
§  563,  the  coefficients  of  like  powers  of  x  in  the  two  members 
are  equal. 

That  is,  5^  +  35  =  19, 

and  2^-     J5  =  l.   • 

Solving  these  equations,  we  obtain  A  =  2  and  B  =  3. 
Substituting  in  (1),  we  have 

19x  +  l         _      2      _^      3 


(3a;-l)(5a5  +  2)      3a;-l      5x  +  2 

The  result  may  be  verified  by  finding  the  sum  of  the  partial 
fractions. 

'  2.   Separate ^-t — —  into  partial  fractions. 

^  X  —  X^  —  3/ 

The  factors  of  2x  —  x^  —  x^  are  x,  1  —  x,  and  2  +  a;. 

(a  4.V  a; +  4  A  ,      B     ,      C 

Assume  then -^ =  -  +  z h  tt-, — 

I  2x  —  Qif  —  x^      X      1  —  x     2  +  x 

Clearing  of  fractions,  we  have 

x-\-^  =  A(l-x)(2-j-x)-{-Bx(2  +  x)-{-Ox(l-x). 

This  equation,  being  satisfied  by  every  value  of  x,  is  satisfied 
when  x  =  0. 

/f    IV 


384  ADVANCED  COURSE  IN   ALGEBRA 

Putting  ic  =  0,  we  have  4  =  2  A,  or  A  =  2. 

Again,  the  equation  is  satisfied  when  x  =  l. 

Putting  a;  =  1,  we  have  5  =  3  B,  or  B  =  — 

The  equation  is'  also  satisfied  when  x  =  —  2. 

Putting  x  =  —  2,  we  have  2  =  —  6(7,  or  C= 

o 

5  1 

mt.                     ^  +  4           2        3  3 

Then,  o^      ^     ^  =  "  +  q z  + 


a;     3(1-0?)      3(2  +  a;) 

The  student  should  compare  the  above  method  of  finding  A  and  B  with 
that  used  in  Ex.  1. 

568.   Case  II.     All  the  factors  of  the  denominator  equal. 

^2 11  £c  -I-  26 

Let  it  be  required  to  separate   — ^ — •  into  partial 

fractions.  ^^  ~  ^) 

Substituting  ^  +  3  for  x,  the  fraction  becomes 

(y  +  3)^-ll(y  +  3)  +  26^i/^-5y  +  2^1      5  ^  2 

f  f  y    y^    t 

Replacing  2/  by  aj  —  3,  the  result  takes  the  form 

1  5  2 

a;_3      {x-ZY     {x-Zy^' 

This  shows  that  the  given  fraction  can  be  expressed  as  the 
sum  of  three  partial  fractions,  whose  numerators  are  indepen- 
dent of  X,  and  whose  denominators  are  the  powers  of  a;  —  3 
beginning  with  the  first  and  ending  with  the  third. 

Similar  considerations  hold  with  respect  to  any  example 
under  Case  II;  the  number  of  partial  fractions  in  any  case 
being  the  same  as  the  number  of  equal  factors  in  the  denomi- 
nator of  the  given  fraction. 

^x.   Separate — — ^  into  partial  fractions. 

\0  X  -\-  Oj 


UNDETERMINED   COEFFICIENTS  385 

In  accordance  with  the  above  principle,  we  assume  the  given 
fraction  equal  to  the  sum  of  two  partial  fractions,  whose  de- 
nominators are  the  powers  of  3  a?  +  5  beginning  with  the  first 
and  ending  with  the  second. 

That  IS,  — — ^'^— -  = -f 


(3i»  +  5)2     3x  +  b     (3a;  +  5)2 
Clearing  of  fractions,  6a?-f-5  =  ^(3a;  +  5)+5 

JEquating  coefficients  of  like  powers  of  x, 

SA  =  6, 

and  5^4-^  =  5. 

Solving  these  equations,  A  =  2  and  B  =  —  5. 

Whence  6a;4-5   ^2  o 

(3x-j-5y     3a;  +  5      (3x-{-5y 

569.   Case  III.    Some  of  the  factors  of  the  denominator  equal. 

Ex.   Separate  — - — — —  into  partial  fractions. 

x{x-{-  ly 

The  method  in  Case  III  is  a  combination  of  the  methods  of 
Cases  I  and  II ;  we  assume 


xix^lf      X      x-\-l      (x+Vf     {x  +  1) 
Clearing  of  fractions, 

3x +  2  =:  A(x -^If +  Bx(x^iy+ Cx{x -{-!)+ Dx 
=  (^ +  5)i»3  +  (3^  +  2  jB  +  (7)a^ 

-^{3A-\-B-\~C+D)x^-A. 
Equating  coefficients  of  like  powers  of  x, 

A  +  B^O, 

3^  +  2^4-0  =  0, 

3J.  +  J3+C  +  7)  =  3, 

and  A  =  2. 


386  ADVANCED   COURSE   IN   ALGEBRA 

Solving,  we  have  A  =  2,  B  =  -2,  C=-2,  and  D  =  l. 
Substituting  in  (1), 

3x  +  2   _2         2  2  1 


x{x-^iy     X     x  +  1      (a; +  1)2      {x  +  Vf 

The  following  general  rule  for  Case  III  will  be  found  convenient : 

A  fraction  of  the  form  should  be  assumed 

equal  to  (x  +  a){x+hy.-ix  +  my...     . 

A  B  E  F  T^ 

+  — ^+--+-^^^+  .  ..  +  •••  + 


x-{-  a     x-\-h  X  -\-  m      {x  +  my  (x  +  my 

Single  factors  like  x -\-  a  and  x  -i-b  having  single  partial  fractions  cor- 
responding, arranged  as  in  Case  I ;  and  repeated  factors  like  (x  +  my 
having  r  partial  fractions  corresponding,  arranged  as  in  Case  II. 

EXERCISE  87 

Separate  the  following  into  partial  fractions  : 

75  g    5x2  +  4a;-l 


1. 

Sx-6 

5 

4x2-9 

2. 

19  a;  -  30 
5  x2  -  6  a; 

6.    ' 

3. 

2  a; -10 

(2x-3)2 

7.    • 

4. 

2  a;2  +  15  a:  +  34 

8 

{x  +  5y 

^ 

62  a;  -  38 

a:3-25x  (5x  +  2)8 

9a;2-15a;  ^^  5 ax2  -  2  a'^x -Sa^ 

(Sx-iy  '  x^  +  Sax^-4a^x' 

43x  +  6  J,  J  4x2-22x4-63 


6x2  +  5x-6  x(x-3)2 

8x2  j2         42-27X 


8  -  14  X  +  5  x2 
2x^+  19x2-17x  +  12_ 
(2x-l)(12x2-x-6)  ""  x4-|-4x3 

16    a;8  -  X  -  4  ^g      18x-4V2  ^^       9x2-9x-18 


(2x- 

-3)3 

14.    ■ 

18  X- 

-4V2 

(x-l)4  9x2-f6x-l  (x2  -  2  x)  (x2  -  9) 

18      14x2+lla;-f29  21.   ^x^  -  24  xMi20x--4 


19. 


(3x-l)(2x-h3)2  (3x-2)* 

x3  +  14x2-1- 9  X  oo    2  x3 -f  12  x2 -h  12x+i 


(x4-2)4  x(x+l)(x-f-2)2 

2Q    x3-3x2-7x-f4  23    19  x  -  32 

x2(x-l)2        '  '    (2x-3)(8x2- lOx-3)' 

570.  If  the  degree  of  the  numerator  is  equal  to,  or  greater 
than,  that  of  the  denominator,  the  preceding  methods  are 
inapplicable. 


UNDETERMINED   COEFFICIENTS  387 

In  such  a  case,  we  divide  the  numerator  by  the  denominator 
until  a  remainder  is  obtained  which  is  of  a  lower  degree  than 
the  denominator. 

Ex.    Separate  — into  an  integral  expression  and 

partial  fractions.  ~ 

We  have,  by  division,  — ^ ^^—  =x  —  2-\ .       (1) 

x^  —  x  x^—x 

2ic  —  1 

We  can  now  separate  into  partial  fractions  by  the 

Ou    —  X      -I  Q 

method  of  Case  I;   the  result  is -• 

X       X—  1 

Substituting  in  (1),  — ^ ^—  =  x  —  2-\ . 

Olf  —  X  X        X  —  1 

Another  way  to  solve  the  above  example  is  to  combine  the  methods  of 
§§  664  and  667,  and  assume  the  given  fraction  equal  to 

C  .      D 


Ax  +  B  + 


X        X  —  1 


EXERCISE  88 

Separate  the  following  into  integral  expressions  and  partial  fractions  : 

J    9  a;3  +  21  x2  +  22  a;  -  17  ^    2x^-2  x^ -3  x"^-]-! 

(a:  +  2)(3a;-l)       *  *  x^  -  x^ 

2    4  a;3  -  26  a;2  +  20  X  -  1  _  ^    6x^  +  6x^-2x^  +  6x  +  l 

(X-2)3  *  •  ic2(^X  +  l)3 

5  x6  +  15  x5  4-  3  x*  -  14  x2  -  18 


5. 


X*  +  3  x3 


571.  If  the  denominator  of  a  fraction  can  be  resolved  into 
factors  partly  of  the  first  and  partly  of  the  second,  or  all  of  the 
second  degree,  in  x,  and  the  numerator  is  of  a  lower  degree 
than  the  denominator,  the  Theorem  of  Undetermined  Coeffi- 
cients enables  us  to  express  the  given  fraction  as  the  sum  of 
two  or  more  partial  fractions,  whose  denominators  are  factors 
of  the  given  denominator,  and  whose  numerators  are  inde- 
pendent of  X  in  the  case  of  fractions  corresponding  to  factors 
of  the  first  degree,  and  of  the  form  Ax-\-B  in  the  case  of 
fractions  corresponding  to  factors  of  the  second  degree. 


388  ADVANCf:D  COURSE  IN  ALGEBRA 

The  only  exceptions  occur  when  the  factors  of  the  denorai- 
nator  are  of  the  second  degree  and  all  equal. 

Ex.   Separate  — — -  into  partial  fractions. 

The  factors  of  the  denominator  are  x-\-l  and  x'  —  x-\-l. 

Assume  then     ^^—  =  -^  +   ^^  +  ^    .  (1) 

Clearing  of  fractions,  we  have 

l  =  A{x'-x  +  l)  +  {Bx  +  C){x^-l) 
=  {A  +  B)x'  +  {-A  +  B  +  C)x  +  A-^a 
Equating  coefficients  of  like  powers  of  a?, 

A  +  B  =  0, 

■^^  +  5+0  =  0, 

and  '  A-\-G=l. 

1  1  2 

Solving  these  equations,  ^  =  q?    -^  =  ~q'   ^^^  ^~q' 

o  o  o 

Substituting  in  (1),  ~ 


4-1     3(a)  +  l)     Six'-x  +  l) 


EXERCISE   89 

Separate  the  following  into  partial  fractions 


1. 

2. 
3. 

6  x2  +  5  X  +  2 
5  a;2  +  a^  -  14 

4. 
5. 
6. 

4:X^~6x^  +  ex-\-S 

x*-l 
22X-6 

(3x  +  l)(a:2_x  +  3) 

3  a:2  -  ic  +  25 
2  a^3  _  5  a;2  +  4  a;  _  10 

8  a;3  -  27 
4x^+lSx  +  S 

ic^  +  3  a;2  +  4 

572.  We  will  now  show  how  to  find  an  expression  for  the 
nth  term  of  the  expansion  of  a  fraction  in  ascending  powers  of 
Xy  when  the  denominator  can  be  resolved  into  binomial  factors 
of  the  first  or  second  degree  in  x. 

Ex.   Find  the  nth  term  in  the  expansion  of  the  fraction 

l  +  7a;         .  ,.  p 

:, ~ -—:  m  ascending  powers  of  x. 

l-j-2  x^3x^ 


UNDETERMINED  COEFFICIENTS  889 

Separating  the  fraction  into  partial  fractions  by  the  method 
of  §  667,  we  have 

l  +  7aj  2  1 


l  +  2a;-3a^     1-x     l  +  3a? 


(1) 


By  division,   -^  =  2(l  +  x-{-a^-{-a^+ -"),  (2) 

J.  ~~~  35 

and  \     =i_3a;  +  3V-3V+-.  (3) 

J-  "T"  O  3/ 

The  nth  term  of  (2)  is  2a;'*~^,  and  the  nth.  term  of  (3)  is 
(_3)»-ia;«-i. 

Subtracting,  the  nth  term  of  (1)  is 

2  x^-i  -  (-  3)"-ia;«-S  or  [2  -  (-  3)"-i] a;"-^ 

ITie  above  may  be  used  as  a  formula  to  obtain  the  successive  terms  of 
the  expansion. 

If  w  =  1,  the  expression  becomes  (2  —  l)x'',  or  1. 

li  n  =  2,  the  expression  becomes  (2  +  3)aj,  or  5  x. 

If  n  =  3,  the  expression  becomes  (2  —  9)x^,  or  —  7  x^  j  etc. 

Then,  l±l^—=l  +  6x-7x^  +  ".. 

1  +  2  X  -  3  x2 

EXERCISE  90 

In  each  of  the  following,  find  the  nth  term  of  the  expansion  of  the 
fraction  in  ascending  powers  of  x : 

1  3  ^         5  a; -12 


l-x-2a:2  ex2  +  5x-6 

2  2x-l  g  7  4-  4  x  -  x2 

'   1  -  9  X  +  20  a?*  '   6  +  3  X  -  4  x2  _  2  x3* 

8         6 +  11  a;  g      1  +  5  x  -f  6  x^^ 
'   2  +  5x-3x2'  •    (l_a;)(l+x2)* 

In  Exs.  5  and  6,  it  should  be  observed  that  there  are  two  forms  for  the 
nth  term  according  aa  n  is  even  or  odd. 


REVERSION  OP  SERIES 

573.  To  revert  a  given  series  y  =  a-\- baf* -f  ca;"  +  •••  is  to  ex- 
press X  in  the  form  of  a  series  proceeding  in  ascending  powers 
of  2/. 


390 


ADVANCED   COURSE   IN   ALGEBRA 


Ex.     Eevert  the  series  2/  =  2aj  —  3x^-1-40^  —  5x*H — 

Assume,  x  =  Ay-{-  By^  -f  Cy^  +  Dy^  H . 

Substituting  in  this  the  given  value  of  y, 

+  5(4a;2  +  9ic*_12a^  +  16a;*4-...) 

+  0(8  a;3  -  36  o;^ +...)+ J9(16  a;^ +•••)  + ■ 


a;2+    4^ 

aj3-    5^ 

-12  5 

+  25  5 

+   8(7 

-36(7 

+  16i) 

That  is,  a;  =  2^a;-3^ 
+  45 


Equating  coefficients  of  like  powers  of  Xy 

2A  =  1 
-SA+4B=0 
4^-125  +  8(7=0 
-5^  +  255-36  (7+16  2)  =  0;  etc. 


a7^+.. 


Solving,  A  =  ^,  B  =  j,  C: 
Substituting  in  (1),  x: 


16' 


i>  =  ^,etc. 


1  3     2,5_, 

2^ "^8^  '^W 


128^ 


(1) 


If  the  even  powers  of  x  are  wanting  in  the  given  series,  the 
operation  may  be  abridged  by  assuming  x  equal  to  a  series  con- 
taining only  the  odd  powers  of  y. 

Thus,  to  revert  the  series  2/  =  a;  —  o^  +  a^  —  a;^  H ,  we  assume 

x  =  Ay  +  Bf+Cf-hDf  +  '':. 

If  the  odd  powers  of  x  are  wanting  in  the  given  series,  we 
substitute  t  for  x^,  and  revert  the  series,  expressing  t  in  ascend- 
ing powers  of  2/ ;  by  taking  the  square  root  of  this  result,  x 
itself  may  be  expressed  in  ascending  powers  of  y. 

If  the  first  term  of  the  given  series  is  independent  of  x,  it  is 
impossible  to  express  x  in  ascending  powers  of  y,  though  it  is 
possible  to  express  it  in  the  form  of  a  series  whose  terms  are 
functions  of  y. 


UNDETERMINED  COEFFICIENTS  391 

Thus,  let  it  be  required  to  revert  the  series 

^  [2      |3 

The  series  may  be  written  2/  —  1  =  a?  +  —  +  —  +  •••. 

[2     [3 

We  then  assume 

05  =  ^(2/ -  1)  +  J5(2/ - 1)2  +  (7(2/ - 1)3  +  Z>(2/ - 1)*  + -. 
Proceeding  as  before,  we  find 

aj  =  (2^-l)-|(2/-l)2  +  |(2/-l)«-J(2/-l)*+.... 


EXERCISE  91 

Revert  each  of  the  following  to  four  terms  : 

1.  ^/zzzXTf  3x2  +  5a;3  +  7  a:*  +  •-. 

2.  ?/ =  x- 2x2 +  3x3-4x4+ •••• 

/>»2         /v<3         />*4 

^  2       3       4 

4.  ?/  :=  1  +  2  X  +  5  x2  +  8  x3  +  11  X*  + 

/>•        0*2        'vS        />»4 

5.  ^  =  - -  —  +  —  --+ .... 

^2      4^6       8 
«     ,,      X  ,  x2  ,  x^  . 


7.   y  =  2x-4a;3  +  6x5-8x''  + 

?!  +  ^_i-?! 

13      [6     II 


a:+.^  +  g+5!  + 


392  ADVANCED   COURSE  IN  ALGEBRA 


XXVIII.    THE  BINOMIAL  THEOREM 

ANY  RATIONAL  EXPONENT 

574.  It  was  proved  in  §  285  that,  if  ?i  is  a  positive  integer, 
(l  +  ^)-  =  l  +  na^  +  ^^^^^^  +  "^^-l^,^^-^>x3+  -...     (1) 

In  this  case,  the  second  member  is  a  finite  series  of  n-\-l 
terms. 

If  n  is  a  negative  integer,  or  a  positive  or  negative  fraction, 
the  series  in  the  second  member  becomes  infinite ;  for  no  one 
of  the  expressions  n  —  1,  n  —  2,  etc.,  can  equal  zero. 

We  will  now  prove  that  in  this  case  the  series  gives  the  value 
of  (1  -f  xy,  provided  it  is  convergent;  this  we  know  to  be  the  case 
when  X  is  numerically  <  1  (§  559). 

575.  Proof  of  the  Binomial  Theorem  for  Any  Rational  Exponent. 

If  m  and  n  are  positive  integers, 

(l+x)-^=l  +  mx-\-  ^'(^  ~  ^^  a^  +  -,  (2) 

If 

and  (l+xy  =  l  +  nx-{-'*^^'\'~^^x^+  ....  (3) 

But,  (1  +  x)"^  X  (1  +  xy  =  (1  4-  a?)'""^".     • 

Then  the  product  of  the  series  in  the  second  members  of  (2) 
and  (3)  must  equal  the  expanded  form  of  (1  -f  0?)"^+" ;  that  is, 

=  l+(m  +  n>+(^  +  ^>^,^  +  ^~^V+  ....  (4) 

We  proved  the  above  result  on  the  hypothesis  that  both  m 
and  n  were  positive  integers. 


THE  BINOMIAL   THEOREM  898 

But  the  form  of  the  product  will  evidently  be  the  same  what- 
ever the  values  of  m  and  n. 

Therefore,  (4)  holds  for  all  rational  values  of  m  and  n,  pro- 
vided X  is  numerically  <1;  for  in  this  case  each  series  is 
convergent. 

(We  assume  that  the  product  of  two  convergent  series  is  convergent.) 

Now  let  the  symbol /(m)  stand  for  the' series 
^    ,  ,  m  (m  —  1)   o  , 

for  any  rational  value  of  m. 

Then,  if  m  is  a  positive  integer,  /(m)  =  (1  +  x)"^,  (5) 

Then,  by  §  251,  (4)  may  be  written 

■  /(m)x/(r0=/(m+7i),  (6) 

which  holds  for  all  rational  values  of  m  and  n. 

Then  by  (6),  if  p  is  also  a  rational  number, 
f(m)  X  f(n)  X  /(p)  =  [/(m  +  n)]  x /(p)  =/(m  +  n  H-p)  ;  etc. 
Thus,  /(m)  x/(n)  x/(p)  X  •••  to  r  factors 

=f(m  +  w  +p  +  •  •  •  to  r  terms).  (7) 

ISTow  let  m,  ti,  p,  •••  be  each  equal  to  -,  where  q  and  r  are 
positive  integers;  then  (7)  becomes 

[/(f)]=/(fx.)=/(,). 

But  since  ^  is  a  positive  integer,  f(q)  =(l-\-xy. 

Then,  [fg)j  =  (l  +  a!)'. 

Taking  the  rth  root  of  both  members, 

'         /N  -f--0 

(l  +  xX=/(?j  =  l  +  fx+^l^^a^  +  ...;  (8) 

which  proves  the  theorem  for  a  positive  fractional  exponent. 
The  result  is  proved  only  for  the  case  where  x  is  numerically  <  1. 


394  ADVANCED  COURSE  IN  ALGEBRA 

Again,  in  (6),  let  m  =  —  w,  where  n  is  a  positive  integer  or 
a  positive  fraction. 

Then,  /(-  71)  xf(n)  =/(-  n  +  n)  =/(0). 

But/(0)  stands  for  {1  +  xf,  which  equals  1  (§  359). 

Therefore,  /(  -  n)  =  -J-  • 

Now  since  w  is  a  positive  integer  or  a  positive  fraction,  it 
follows  from  (5)  and  (8)  that/(w)  =  (1  +  x)\ 

Whence,  f(—n)  = 

'  ^v        ;       (l  +  iC)'* 

Then, 

which  proves  the  theorem  for  a  negative  integral  or  negative 
fractional  exponent. 

The  result  is  proved  only  in  the  case  where  x  is  numerically  <  1. 

576.   Putting  -  for  x,  in  (1),  §  574, 

\       aj  a  \2_       a^  ^  ^ 

Multiplying  each  term  by  a%  we  have 

(a  +  xY  ==  a"  +  7ia''-^x  +  ^^^  ""-*-)  a'*- V  +  . . . .  (lO) 

We  know  that  the  second  member  of  (9)  is  convergent  when- 

-  is  numerically  <  1. 
a 

Hence,  the  second  member  of  (10)  is  convergent  when  x  is 
numerically  <  a. 

If  x  is  numerically  >  a,  we  can  expand  (a  +  xy  in  ascending 
powers  of  a ;  thus, 

(a;  +  ay  =  aj"  +  nx^'-^a  -f  ^'  ^^  ~  -^^  a;"- V  +  •  •  • ; 


and  this  series  will  be  convergent,  since  a  is  numerically  <  x, 


THE   BINOMIAL   THEOREM  395 

577.  Examples. 

In  expanding  expressions  by  the  Binomial  Theorem  when 
the  exponent  is  fractional  or  negative,  the  exponents  and 
coefficients  of  the  terms  may  be  found  by  the  laws  of  §  285, 
which  hold  for  all  rational  values  of  the  exponent. 

1.  Expand  (a  +  x)^  to  five  terms. 

The  exponent  of  a  in  the  first  term  is  -,  and  decreases  by  1 
in  each  succeeding  term. 

The  exponent  of  x  in  the  second  term  is  1,  and  increases  by 
1  in  each  succeeding  term. 

The  coefficient  of  the  first  term  is  1 ;  of  the  second  term,  -. 

2  1 

Multiplying  -,  the  coefficient  of  the  second  term,  by , 

o  o 

the  exponent  of  a  in  that  term,  and  dividing  the  product  by 

the  exponent  of  x  increased  by  1,  or  2,  we  have  —  ^  as  the 
coefficient  of  the  third  term  ;  and  so  on. 
Then, 

(a  +  x)^  =  at  -f  ?a-^ a;  -  ^  oT^  x'  +  ^orix^-  ^oT^^  ic^  +  •... 
^  ^  3-9  81  243 

It  follows  from  §  576  that  the  series  expresses  the  value  of  (a  +  xy 
only  when  x  is  numerically  <  a.  ^ 

If  X  is  numerically  >  a,  (a  +  xY  is  equal  to  i   , 

x^ +  -x~^  a --x' ^  a^ -{■ —x'"^  a^ . 

3  9  81 

2.  Expand  (1  +  2a;~^)-2  to  five  terms.  i. 
Enclosing  2x~^  in  parentheses,  we  have 

(1+2  a;-^)-2  =  [1  +  (2  a;-^)]-^ 

=  1-2-  2  . 1-3 .  (2«-^)  +3.1-^.  {2x-^Y 

-  4  . 1-^ .  (2 x-'^f  4-  5  .  l-«  •  (2 x-^y . 

=  1  _4  aj-i  4_  12  x-^  -  32  ic"^  +  80  x-^  -\ . 

By  writing  the  exponents  of  1,  in  expanding  [1  +  (2a;~^)]-2,  we  can 
make  use  of  the  fifth  law  of  §  285. 


396  ADVANCED  COURSE   IN  ALGEBRA 

The  series  expresses  the  value  of    (1+2  a;~2)-2  only  when  2  x"^   is 

numerically  <  1,  or  x~^  numerically  <  -• 
1  2 

That  is,  when  x^  is  numerically  >  2,  or  when  x  is  positive  and  >  4. 

3.   Expand    ^  —  to  four  terms. 

Enclosing  a~^  and  —  3  tc^  in  parentheses,  we  have 
3,      ^  = ^^  =  [(«-')  +  (- 3  x*)]-* 

=  (a-')'*  -  \  («-'r^  (-  3  ^')  + 1  (a->)'^  (-  3  x^f 
-|f(«-T^(- 3. *)»  +  ••• 

=  a^  +  aM  +  2  a^a;^  +  ^a^x  -\ . 

The  result  expresses  the  value  of  the  given  fraction  only  when  Zx^ 

is  numerically  <  a-i  ;  if  3  x"3  is  numerically  >  a-i,  the  fraction  can  be 
expanded  in  ascending  powers  of  a-i. 

EXERCISE  92 

Expand  each  of  the  following  to  five  terms : 
1.    {a  +  xy. 


2. 

(l  +  ^)-«. 

8. 

(1  -  xy. 

4. 

^a-x. 

5 

1 

(a  +  6)8 

6. 

1 

7.    {x^-2yy.  12.    — ^— i ^. 

(m^  -  2  n~^Y 


8.  H+V 

9.    ia^-2xh-k  ^'     ""' 


13. 

\h     a) 

10.  1  14.    {x~^-Zyh'^. 


15.    f— 1 ^^2\l 

v^n^  11.    ^[(a:-3  +  6?/0)7].  \16\/^  / 

678.  The  formula  for  the  rth  term  of  {a  +  xf  (§  287)  holds 
for  fractional  or  negative  values  of  n,  since  it  was  derived  from 
an  expansion  which  has  been  proved  to  hold  for  all  rational 
values  of  the  exponent. 


THE   BINOMIAL   THEOREM  397 

Ex.   Find  the  7th  term  of  (a  -  3  x'^y^. 
Enclosing  —  3  x~^  in  parentheses,  we  have 

(a  -  3  x-^y^  =  [a  4-  (-  3  x-^)Y'^. 
The  exponent  of  (—  3  x~^)  is  7  —  1,  or  6. 

1  19 

The  exponent  of  a  is 6,  or • 

o  o 

The  first  factor  of  the  numerator  is  — -,  and  the  last  factor 

19  ,  -,  16  ^ 

-^  +  l,or--. 

The  last  factor  of  the  denominator  is  6. 
Hence,  the  7th  term 

_1  _4  _7  _10  _13  _16 
3*3*3*      3  *      3  *      3 


1.2.3.4.5.6 

728   -Jta,Q6  -9N      728   - 


a-¥(_3a.-f)6 


hr 


EXERCISE  93 

Find  the 

1.  6th  term  of  {a  +  xy.  7.  7tli  term  of  (a*  -  sc^) 

2.  5th  term  of  (a  +  6)""^.  8.  10th  term  of  ^ 

3.  12th  term  of  (1  -  a;) -5.  ^^  ^  "*^^ 

4.  7th  term  of  (^-1  +  2  2/^)-2.  9.  8th  term  of  (m^  -  2  Q-J 

5.  9th  term  of  (a  +  2  x)^.  ^^^    ^^^^  ^^^"^  ^^  ^^^  "  ^^'• 

1  11.   6th  term  of  {a^  -  4  6-2)  ^. 

6.  5th  term  of  ^  1  _2 

V(l  -  a:)5  12.    8th  term  of  (x'^  +  3  y~^y^. 

13.  Term  involving  xr^^  in  f  x  Vw^  +  -rrz  I     * 

_9  1  _2  '9 

14.  Term  involving  a  -^  in  (a^  —  4  &  ^)*. 

579.   Extraction  of  Roots. 

The  Binomial  Theorem  may  sometimes  be  used  to  find  the 
approximate  root  of  a  number,  which  is  not  a  perfect  power  of 
the  same  degree  as  the  index  of  the  root. 


398  ADVANCED   COURSE  IN   ALGEBRA 

Ex.     Find  V25  approximately  to  five  places  of  decimals. 

The  nearest  perfect  cube  to  25  is  27. 

We  have  ■^/25  =  </27^^  =  [(3^)  +  (- 2)]3 

=  (33)i  +  |(33)-t(_2)-l(33)-t(-2y 

+  |;(30"^(-2)^-- 
o         2  4  40 


3  .  32     9.3^     81  .  3« 

Expressing  each  fraction  approximately  to  the  nearest  fifth 
decimal  place,  we  have 

</25  =  3  -  .07407  -  .00183  -  .00008 =  2.92402. 

In  any  case,  we  separate  the  number  into  two  parts,  the  "first  of  which 
is  the  next  less  or  next  greater  perfect  power  of  the  same  degree  as  the 
required  root ;  choice  being  made  of  the  one  which  makes  the  smaller 
second  term,  provided  the  series  is  convergent  (§  576). 

Thus,  in  finding  ^/Vf,  we  should  take  (S^  -  10)3,  and  not  (23  +  9)^  ; 
for  although  the  latter  has  the  smaller  second  term,  the  series  is  not  con- 
vergent (§  576).  

If  the  ratio  of  the  second  term  of  the  binomial  to  the  first  is  a  small 
proper  fraction,  the  terms  of  the  expansion  diminish  rapidly ;  but  if  this 
ratio  is  but  little  less  than  1 ,  it  requires  a  great  many  terms  to  insure  any 
degree  of  accuracy. 

EXERCISE  94 

Find  the  approximate  value  of  each  of  the  following  to  five  places  of 
decimals  : 

1.  VsT.  3.    v^iSO.  5.    ^/TO.  , 

2.  Vll8.  •  4.    v^l8.  6.    \/223. 


LOGARITHMS 


XXIX.    LOGARITHMS 

580.  Any  positive  real  number,  m,  can  be  expressed  in  the 
form  a*,  where  a  is  any  positive  real  number  except  unity. 

If  a*  ==  m,  X  is  called  the  Logarithm  of  m  to  the  Base  a;  a 
relation  which  is  expressed 

X  —  log^m. 

A  negative  number  is  not  considered  as  having  a  logarithm. 

581.  The  Common  System. 

Logarithms  *  of  numbers  to  the  base  10  are  called  Common 
Logarithms,  and  collectively  form  the  Common  System. 

They  are  the  only  ones  used  for  numerical  computation. 

It  is  customary,  in  writing  common  logarithms,  to  omit  the 
subscript  10  which  indicates  the  base ;  thus, 

logio  13  is  written  simply  log  13. 

Any  positive  real  number,  except  unity,  may  be  taken  as  the  base  of  a 
system  of  logarithms. 

582.  By  §§  359  and  360, 

^Jf      10«  =  1,  10-1  =  .1, 

1^/^              10^  =  10,  10-2  =  .01, 

W  =  100,  10-3  ^  001^  etc. 
Whence,  by  the  definition  of  §  581, 

logl  =  0,  log.l=-l  =  9-10, 

log  10  =  1,  log..01  =-  2  =  8  -  10, 

log  100  =  2,  log  .001  =  -  3  =  7  -  10,  etc. 

The  second  form  for  log  .1,  log  .01,  etc.,  is  preferable  in  practice. 
Where  no  base  is  expressed,  the  base  10  is  understood. 

/    It  is  evident  from  the  above  that  the  common  logarithm  of 
J  a  number  greater  than  1  is  positive,  and  of  a  number  between^ 
I  0  and  1  negative. 


400       ADVANCED  COURSE  IN  ALGEBRA 

If  a  number  is  not  an  exact  power  of  10,  its  common  loga- 
rithm can  only  be  expressed  approximately ;  the  integral  part 
of  the  logarithm  is  called  the  characteristic^  and  the  decimal 
part  the  mantissa. 

For  example,  log  13  =  1.1139. 

Here,  the  characteristic  is  1,  and  the  mantissa  .1139. 

A  negative  logarithm  is  always  expressed  with  a  positive 
mantissa,  which  is  done  by  adding  and  subtracting  10. 

Thus,  the  negative  logarithm  —  2.5863  is  written 

(10  -  2.5863)  - 10,  or  7.4137  -  10. 
In  this  case,  7  —  10  is  the  characteristic. 

The  negative  logarithm  7.4137  —  10  is  sometimes  written  3.4137  ;  the- 
negative  sign  over  tlie  cliaracteristic  showing  that  it  alone  is  negative,  the 
mantissa  being  always  positive. 

583.  If  a  number  has  five  places  in  its  integral  part,  it  lies 
between  10'^  and  10^;  and  hence  its  common  logarithm  lies 
between  4  and  5. 

Therefore,  the  characteristic  of  its  logarithm  is  4. 

In  general,  if  a  number  has  n  places  in  its  integral  part,  it 
lies  between  10""^  and  10** ;  and  the  characteristic  of  its  loga- 
rithm is  71  —  1. 

h  Hence,  the  characteristic  of  the  logarithm  of  a  number  greojter 
jlj  than  1  is  1  less  than  the  number  of  places  to  the  left  of  the  deci- 
»'  mal  point. 

584.  If  a  decimal  has  three  ciphers  between  its  decimal 
point  and  first  significant  figure,  it  lies  between  10~^  and  10~^ ; 
and  hence  its  common  logarithm  lies  between  6  —  10  and 
7-10. 

Therefore,  the  characteristic  of  its  logarithm  is  6  — 10. 

In  general,  if  a  decimal  has  n  ciphers  between  its  decimal 
point  and  first  significant  figure,  it  lies  between  10~^""*"^^  and 
10~";  and  its  common  logarithm  lies  between  —(n  -f  1)  and  —  n, 
or  (9  -  n)  - 10  and  (10  -  n)  -  10. 

That  is,  the  characteristic  of  its  logarithm  is  (9  —  n)— 10. 


LOGARITHMS  401 

Hence,  to  find  the  characteristic  of  the  logarithm  of  a  number 
betiveen  zero  and  1,  subtract  the  number  of  ciphers  between  the 
decirr^al  point  and  first  significant  figure  from  9,  writing  — 10 
after  the  mantissa. 


SRTIE3  OF  T2 

585.  In  any  system,  the  logarithm  of  uiiity  is  zero^ 
For  by  §  359,  a"  =  1 ;   whence,  by  §  580,  log„  1  =  0. 

586.  In  any  system,  the  logarithm  of  the  base  is  umtVy 
Eor  a^  =  a',  whence,  log^a  =  1. 

^   587.   In   any  system  whose  base  is  greater  than  unity,  the 
logarithm  of  zero  is  minus  infinity. 

For  if  a  is  >1,  a—  =  4  =  -  =  0  (§  248). 

a       00 

Whence,  by  §  580,  log„0  =  -  oo. 

The  above  result  must  be  interpreted  as  follows  : 

If,  in  any  system  whose  base  is  greater  than  unity,  a  number  approaches 
the  limit  0,  its  logarithm  is  negative,  and  increases  without  limit  in  abso- 
lute value.     (Compare  §  248.) 

588.  In  any  system  whose  base  is  less  than  unity,  the  logarithm 
of  zero  is  infinity. 

For  if  a  is  <1,  a"^  =  0 ;  whence,  log^O  =  oo. 

This  means  that  if,  in  any  system  whose  base  is  less  than  unity,  a 
number  approaches  the  limit  0,  its  logarithm  increases  without  limit. 

589.  In  any  system,  the  logarithm  of  a  product  is  equal  to  the 
sum  of  the  logarithms  of  its  factors.  ^     U        C?/'  '  /" 

Assume  the  equations  a'' =  m,  a^  =  n.    '^ 

Then,  by  §  580,     x  =  log«  m,  y  —  log,,  n,  X'  '^ 

Multiplying  the  assumed  equations, 

a"  X  a^  =  mn,  or  a^^^  =  mn. 

Whence,        log„  mn  =  x-\-y  =  log^  m  +  log^  n. 

In  like  manner,  the  theorem  may  be  proved  for  the  product 
of  any  number  of  factors. 


1^  ADVANCED  COURSE  IN  ALGEBRA 

By  aid  of  the  above  theorem,  the  logarithm  of  any  positive 
integer  may  be  found  when  the  logarithms  of  its  factors  are 
known. 

Ex.   Given  log  2  =  .3010,  and  log  3  =  .4771;   find  log  72. 

log72  =  log(2x2x2x3x3) 

=  log2  +  log2  +  log2  +  log3 +  log3 

=  3xlog2  +  2xlog3       * 

=  .9030  +  .9542  =  1.8572. 

590.  In  any  system,  the  logarithm  of  a  fraction  is  equal  to 
the  logarithm  of  the  numerator  minus  the  logarithm  of  the 
denominator.  ^ 

Assume  the  equations  a""  =  m,  a^  =  n. 

Then,  x  =  log„  m,  y  =  log„  n. 

a^     m>  771 

Dividing  the  assumed  equations,  —  =  — ,  or  a*~^  =  — 

a^      7i'  n 

Whence,  log,,  ~-z=x  —  y  =  log„  m  —  log^  n. 

Ex.    Given  log  2  =  .3010;    find  log  5. 

log  5  =  log  i^  =  log  10  -  log  2  =  1  -  .3010  =  .6990. 

591.  In  any  system,  the  logarithm  of  any  power  of  a  quantity 
is  equal  to  the  logarithm  of  the  quantity  multiplied  by  the  exponent 
of  the  power. 

Assume  the  equation  a^'^m-,   whence,  a?  =  log„m. 

Raising  both  members  of  the  assumed  equation  to  the  ^th 

power, 

(jfPx  _  ^^p .   -whence,  log„  (m^)  =px=p  log„  m. 

592.  In  any  system,  the  logarithm  of  any  root  of  a  quantity 
is  equal  to  the  logarithm  of  the  quantity  divided  by  the  index 
of  the  root. 

For,  log«  -Vm  =  log„  (m^-)  =  -  log„  m  (§  591). 


LOGARITHMS       ^^S  -  403 

593.  Examples. 

1.  Given  log  2  =  .3010;   find  \og2i 

log  2*  =  ^  X  log  2  =  I  X  .3010  =  .5017. 
o  o 

To  multiply  a  logarithm  by  a  fraction,  multiply  first  by  the  numerator, 
and  divide  the  result  by  the  denominator. 

2.  Given  log  3  =.4771;   find  log  ■y'S.^   .  -     J^ -^  ^ 

log -V3  =  1^  =  4^  =  .0596. 
8  8 

3.  Given  log  2  =  .3010,  log  3  =  .4771,  find  log  (2^  x  3^). 
By  §  589, 

log  (2^  X  3^)  =  log  2^  +  log  3*  =  i  log  2  + 1  log  3  =  .6967. 

EXERCISE  95 

Given  log2  =  .3010,  log  3  =  .4771,  log  7  =  .8451,  find  the  logarithm  of: 


1.   84. 

6.  2^. 

11.    V105. 

16. 

18522. 

2.   392. 
3.f. 

7.  ^3. 

8.  5292. 

12.  43I,. 

13.  2807. 

17. 
18. 

3^ 
/18\* 

4.    12^f. 

9     40^ 
14* 

14.    75^. 

19. 

^36 
7* 

5.    76. 

10.  453. 

15.    \/98. 

20. 

(2^  X  15^). 

594.  To 

prove 

tJie  relation 

log^m 

log„m 

' 

Assume  the  equations  a"  =  m,  ¥  =  m.  ^ 

Then,  x  =  log„m,   y  =  log^m.  ^         ^^  ^ 

From  the  assumed  equations,  a'  =  ¥,  or  a"  =  b.  i  Mr^A    "^         ^y^ 

X  X  flf"^ 

Therefore,  log„  6  =  - ,  or  2/  =  ,--^-  ^^  \  v^^-^     ^  ^ 

y  log«6    '>WA)r^\  } 


404  ADVANCED   COURSE   IN   ALGEBRA 

That  is,  logjam  =  -2£«^. 

By  aid  of  this  relation,  if  the  logarithm  of  a  number  m  to 
a  certain  base  a  is  known,  its  logarithm  to  any  other  base  b 
may  be  found  by  dividing  by  the  logarithm  of  b  to  the  base  a. 

595.  To  prove  the  relation 

logj  a  X  log„  6  =  1. 
Putting  m  =  a  in  the  result  of  §  594, 

log.a  =  |-^  =  ^  r§586). 

l0ga&  log,,  6 

Whence,^  log^  a  x  log„  6=1. 

596.  In  the  common  system,  the  maiitissm  of  the  logarithms  of 
numbers  having  the  same  sequence  of  figures  are  equal. 

Suppose,  for  example,  that  log  3.053  =  .4847. 

Then,  log    305.3  =  log(100  x  3.053)  =  log  100  +  log  3.053 

=  2  4-  .4847  =  2.4847 ; 

log  .03053  =  log  (.01  X  3.053)  =  log  .01  +  log  3.053 

=  8  -  10  +  .4847  =  8.4847  -  10 ;  etc. 

In  general,  if  n  is  any  positive  or  negative  integer, 

log  (10"  X  m)  =  n  log  10  +  log  m  =  n  +  log  m. 

But  10**  X  m  is  a  number  which  differs  from  m  only  in  the 
position  of  its  decimal  point,  and  n  +  log  m  is  a  number  having 
the  same  decimal  part  as  log  m. 

Hence,  if  two  numbers  have  the  same  sequence  of  figures, 
the  mantis  see  of  their  logarithms  are  equal. 

For  this  reason,  only  mantissae  are  given,  in  a  table  of 
Common  Logarithms ;  for  to  find  the  logarithm  of  any  number, 
•we  have  only  to  find  the  mantissa  corresponding  to  its  sequence 
of  figures,  and  then  prefix  the  characteristic  in  accordance  with 
the  rules  of  §§  583  and  584. 

This  property  of  logarithms  only  holds  for  the  common  system,  and 
constitutes  its  superiority  over  other  systems  for  numerical  computation. 


LOGARITHMS  405 

Ex.     Given  log  2  =  .3010,  log  3  =  .4771 ;  find  log  .00432. 
log  432  =  log(2^  X  3«)  =  4  log  2  4-  3  log  3  =  2.6353. 
Then,  the  mantissa  of  the  result  is  .6353. 
Whence  by  §  584,  log  .00432  =  7.6353  -  10. 

EXERCISE  96 

Given  log  2  =  .3010,  log  3  =  .4771,  log  7  =  .8451,  find  the  logarithm  of  : 

1.  87.5.  3.   6750.  5.    .0324.  7.    .784. 

2.  2.592.  4.   274.4.  6.    .000175.  8.    .001875. 

USE  OF  THE  TABLE 

597.  The  table  (pages  406  and  407)  gives  the  mantissae  of 
"the  logarithms  of  all  integers  from  100  to  1000,  calculated  to 
four  places  of  decimals. 

598.  To  find  the  logarithm  of  a  number  of  three  figures. 
Look  in  the  column  headed  "  No."  for  the  first  two  signifi- 
cant figures  of  the  given  number. 

Then  the  required  mantissa  will  be  found  in  the  corre- 
sponding horizontal  line,  in  the  vertical  column  headed  by 
the  third  figure  of  the  number. 

Finally,  prefix  the  characteristic  in  accordance  with  the 
rules  of  §§  583  and  584. 

For  example,     log  168  =  2.2253 ; 

log  .344  =  9.5366  -  10 ;  etc.      -  L*^  7^  t  ^^ 

For  a  number  consisting  of  one  or  two  significant  figures, 
the  column  headed  0  may  be  used. 

Thus,  let  it  be  required  to  find  log  83  and  log  9. 

By  §  596,  log  83  has  the  same  mantissa  as  log  830,  and  log  9 
the  same  mantissa  as  log  900. 

Hence,         log  83  =  1.9191,  and  log  9  =  0.9542. 

599.  To  find  the  logarithm  of  a  number  of  more  than  three 
figures. 


406 


ADVANCED   COURSE  IN  ALGEBJIA. 


No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

lO 

oocto 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

II 

0414 

045J" 

0492 

0531 

0569 

0607 

0645 

0682 

•0719 

0755 

12 

079^ 

0828 

0864 

0899 

0934 

0969-- 

1004 

1038 

1072 

iio6' 

13 

"30 

1173 

1206 

1239 

1271 

1303 

1335 

.1367 

1399 

1430 

14 

1461 

1492 

1523 

1553 

1584 

1614 

1644 

1673 

i703_ 

173^ 

15 

1 761 

1790 

1818 

1847 

1875 

1903 

1931 

1959 

1987 

2014 

i6 

2041 

2068 

2095 

2122 

2148 

2175 

2201 

2227 

2253 

2279 

17 

2304 

2330 

2355 

23S0 

2405 

2430 

2455 

2480 

2504 

2529 

i8 

2553 

2577 

2601 

2625 

2648 

2672 

2695 

2718 

2742 

2765 

19 

2788 

2810 

28^3 

2856 

2878 

29CX) 

2923 

2945 

2967 

2989 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

21 

3222 

3243 

3263 

3284 

3304 

3324 

3345 

3365 

3385 

3404 

22 

3424 

3444 

3464 

3483 

3502 

3522 

3541 

3560 

3579 

3598 

23 

3617 

3636 

3838 

385^ 

3692 

3711 

3729 

3747 

3766 

3784 

24 

3802 

3820 

3874 

3892 

3909 

3927 

3945 

3962 

25 

3979 

3997 

4014 

4031 

4048 

4065 

4082 

4099 

4116 

4133 

26 

4150 

4166 

4183 

4200 

4216 

4232 

4249 

4265 

4281 

4298 

27 

4314 

4330 

4346 

4362 

4378 

4393 

4409 

4425 

4440 

4456 

28 

4472 

4487 

4502 

4518 

4533 

4548 

4564 

4579 

4594 

4609 

29 

4624 

4639 

4654 

4669 

4683 

4698 

4713 

4728 

4742 

4757 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

31 

4914 

4928 

4942 

4955 

4969 

4983 

4997 

5011 

5024 

5038 

32 

5051 

5065 

5079 

5092 

5105 

5119 

5132 

5145 

5159 

5172 

33 

5185 

5198 

5211 

5224 

5237 

5250 

5263 

5276 

5289 

5302 

34 

5315 

5328 

5340 

5353 

5366 

5378 

5391 

5403 

5416 

5428 

35 

5441 

5453 

5465 

5478 

5490 

5502 

5514 

5527 

5539 

5551 

36 

5563 

5575 

5587 

5599 

561 1 

5623 

5635 

5647 

5658 

5670 

37 

5682 

5694 

5705 

5717 

5729 

5740 

5752 

5763 

5786 

38 

5798 

5809 

5821 

5832 

5843 

5855 

5866 

llll 

5888 

5899 

39 

591 1 

5922 

5933 

5944 

5955 

5966 

5977 

5999 

6010 

40 

602 1 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

41 

6128 

6138 

6149 

6160 

6170 

6180 

6191 

6201 

6212 

6222 

42 

6232 

6243 

6253 

6263 

6274 

6284 

6294 

6304 

6314 

6325 

43 

6335 

6345 

6355 

6365 

6375 

6385 

6395 

6405 

6415 

6425 

44 

6435 

6444 

6454 

6464 

6474 

6484 

6493 

6503 

6513 

6522 

45 

6532 

6542 

6551 

6561 

6571 

6580 

6590 

6599 

6609 

66i8 

46 

6628 

6637 

6646 

6656 

6665 

6675 

6684 

6693 

6702 

6712 

47 

6721 

6730 

6739 

6749 

6758 

6767 

6776 

6785 

6794 

6803 

48 

6812 

6821 

6830 

6839 

6848 

6857 

6866 

6875 

6884 

6893 

49 

6902 

691 1 

6920 

6928 

6937 

6946 

6955 

6964 

6972 

6981 

SO 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

51 

7076 

7084 

7093. 

7101 

7110 

7118 

7126 

7135 

7H3 

7152 

52 

7160 

7168 

7177 

7185 

7193 

7202 

7210 

7218 

7226 

7235 

53 

7243 

7251 

7259 

7267 

7275 

7284 

7292 

7300 

7308 

7316 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

1.0, 

0 

1 

2 

3 

4 

5 

6 

7 

8 

0 

LOGARITHMS. 


407 


No. 

0 

1 

2 

3 

4 

6 

6 

7 

8 

9 

55 

7404 

7412 

7419 

7427 

7435 

74+3 

7451 

7459 

7466 

7474 

56 

7482 

7490 

7497 

7505 

7513 

7520 

7528 

7536 

7543 

7551 

57 

7559 

7566 

7574 

7582 

7589 

7597 

7604 

7612 

7619 

7627 

58 

7634 

7642 

7649 

7657 

7664 

7672 

7679 

.7686 

7694 

7701 

59 

7709 

7716 

7723 

7731 

773B 

7745 

7752 

7760 

7767 

7774 

60 

7782 

7789 

7796 

7803 

7810 

7818 

7825 

7832 

7839 

7846 

61 

7853 

7860 

7868 

7875 

7882 

7889 

7896 

7903 

7910 

7917 

62 

7924 

7931 

7938 

7945 

7952 

7959 

7966 

7973 

7987 

63 

7993 

8000 

8007 

8014 

8021 

8028 

8035 

8041 

8048 

8055 

64 

8069 

8075 

8082 

8089 

8096 

8102 

8109 

8116 

8122 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

66 

8195 

8202 

8209 

8215 

8222 

8228 

8235 

8241 

8248 

8254 

67 

8261 

8267 

8274 

8280 

8287 

8293 

8299 

8306 

8312 

8319 

68 

8325 

8331 

8338 

8344 

8351 

8357 

8363 

8370 

8376 

8382 

69 

8388 

8395 

8401 

8407 

8414 

8420 

8426 

8432 

8439 

8445 

70 

8451 

8457 

8463 

8470 

8476 

8482 

8488 

8494 

8500 

8506 

71 

8513 

8519 

8525 

8531 

8537 

8543 

8549 

8555 

8561 

8567 

72 

8573 

8579 

8585 

8591 

8597 

8603 

8609 

8615 

8621 

8627 

73 

8633 

8639 

8645 

8651 

8657 

8663 

8669 

8675 

8681 

8686 

74 

8692 

8698 

8704 

8710 

8716 

8722 

8727 

8733 

8739 

8745 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

76 

8808 

8814 

8820 

8825 

8831 

8837 

8842 

8848 

8854 

8859 

77 

8865 

8871 

8876 

8882 

8887 

8893 

8899 

8904 

8910 

8915 

78 

8921 

8927 

8932 

8938 

8943 

8949 

8954 

8960 

8965 

8971 

79 

8976 

8982 

8987 

8993 

8998 

9004 

9009 

9015 

9020 

9025 

80 

9031 

9036 

9042 

9047 

9053 

9058 

9063 

9069 

9074 

9079 

81 

9085 

9090 

9096 

9101 

9106 

9112 

9117 

9122 

9128 

9133 

82 

9138 

9143 

9149 

9154 

9159 

9165 

9170 

9175 

9180 

9186 

83 

9191 

9196 

9201 

9206 

9212 

9217 

9222 

9227 

9232 

9238 

84 

9243 

9248 

9253 

9258 

9263 

9269 

9274 

9279 

9284 

9289 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

86 

9345 

9350 

9355 

9360 

9365 

9370 

9375 

9380 

9385 

9390 

87 

9395 

9400 

9405 

9410 

9415 

9420 

9425 

9430 

9435 

9440 

88 

9445 

9450 

9455 

9460 

9465 

9469 

9474 

9479 

9484 

9489 

89 

9494 

9499 

9504 

9509 

9513 

9518 

9523 

9528 

9533 

9538 

90 

9542 

'9547 

9552 

9557 

9562 

9566 

9571 

9576 

9581 

9586 

91 

9590 

9595 

9600 

9605 

9609 

9614 

9619 

9624 

9628 

9633 

92 

9638 

9643 

9647 

9652 

9657 

9661 

9666 

9671 

9675 

9680 

93 

9685 

9694 

9699 

9703 

9708 

9713 

9717 

9722 

9727 

94 

9731 

9736 

9741 

9745 

9750 

9754 

9759 

9763 

97.68 

9773 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

96 

9823 

9827 

9832 

9836 

9841 

9845 

9850 

9854 

9859 

9863 

97 

9868 

9872 

9877 

9881 

9886 

9890 

9894 

9899 

9903 

9908 

98 

9912 

9917 

9921 

9926 

9930 

9934 

9939 

9943 

9948 

9952 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 
0 

No. 

0 

1 

2 

3 

4 

5 

6 

7 

8 

408  ADVANCED   COURSE   IN   ALGEBRA 

1.   Kequired  the  logarithm  of  32.76. 
We  find  from  the  table,  log  32.7  =  1.5145, 
log32.8  =  1.5159. 

That  is,  an  increase  of  one-tenth  of  a  unit  in  the  number 
produces  an  increase  jif40014  in  the  logarithm. 

Therefore,  a^-rricregtse  of  six-hundredths  of  a  unit  in  the 
number  wjiiproduce  an  increase  of  .6  x  .0014  in  the  logarithm, 
or  .0008/to  the  nearest  fourthdacimaJ  place. 


Hence,      log  32.76  =  1.5145  +  -0008  A,l-5153. 

that  the  differences  of 


eir  corresponding  num- 


The  above  method  is  based  on  the  assumption 
logarithms  are  proportional  to  the  differences  of 

bers  ;  which,  though  not  strictly  accurate,  is  sufficiently  exact  for  practi- 
cal purposes.  


The  difference  between  any  mantissa  in  the  table  and  the  mantissa  of 
the  next  higher  number  of  these  figures,  is /Called  the  Tabular  Difference. 

The  following  rule  is  derived  from  the  above 

Find  from  the  table  the  mantissajj  of  the  first  three  significant 
figures,  and  the  tabular  diifei^em 

Multiply  the  latter  by  the  remaining  figures  of  the  number,  with 
a  decimal  point  before  them. 

Add  the  result  to  the  mantissa  of  the  first  three  figures,  and  pre- 
fix the  proper  characteristic. 

In  finding  the  correction  to  the  nearest  units'  figure,  the  decimal  por- 
tion should  be  omitted,  provided  that  if  it  is  .5,  or  greater  than  .5,  the 
units'  figure  is  increased  by  1 ;  thus  13.26  would  be  taken  as  13,  30.5  as  31, 
and  22.803  as  23.  '^  ' 


i)8.  -    .  ^ 


2.    Find  the  logarithm  of  .0215^ 

Mantissa  215  =  -3324  Tab.  diff.  =21  r 

2  -08 


.3326  Correction  =  1.68  =  2,  nearly. 

The  result  is  8.3326  - 10. 

600.    To  find  the  number  corresponding  to  a  logarithm. 
1.    Required  the  number  whose  logarithm  is  0.6571. 


LOGARITHMS  409 

Find  in  the  table  the  mantissa  6571. 

In  the  corresponding  line,  in  the  column  headed  "No.,"  we 
find  45,  the  first  two  figures  of  the  required  number,  and  at  the 
head  of  the  column  we  find  4,  the  third  figure. 

Since  the  characteristic  is  0,  there  must  be  one  place  to  the 
left  of  the  decimal  point  (§  583). 

Hence,  the  number  corresponding  to  0.6571  is  4.54. 

2.   Eequired  the  number  whose  logarithm  is  1.3934. 

We  find  in  the  table  the  mantissse  3927  and  3945. 

The  numbers  corresponding  to  the  logarithms  1.3927  and 
1.3945  are  24.7  and  24.8,  respectively. 

That  is,  an  increase  of  .0018  in  the  mantissa  produces  an 
increase  of  one-tenth  of  a  unit  in  the  number  corresponding. 

Then,  an  increase  of  .0007  in  the  mantissa  will  produce  an  in- 
crease of  y^^  of  one-tenth  of  a  unit  in  the  number,  or  .04,  nearly. 

Hence,  the  number  corresponding  is  24.7  -f  .04,  or  24.74. 

The  following  rule  is  derived  from  the  above : 

Find  from  the  table  the  next  less  mantissa^he  tho^ee  figures  cor- 
responding, and  the  tabular  difference.  -^     a?^'^         4^ 

Subtract  the  next  less  from  the  f/iven  mWatissu^'  Un^l  divide  the 
remainder  Jjy  the  tabular  difference 

Annex  the  quotient  to  the  first  three  figures  of  the  number,  and 
point  off  the  result. 

The  rules  for  pointing  off  are  the  reverse  of  tJiose  of  §§  583  and  584 : 
I,   7/*  —  10  is  not  written  after  the  mantissa^  add  1  to  the  characteris- 
tic, giving  the  number  of  places  to  the  left  of  the  decimal  point. 

II.  7/"  —  10  is  written  after  the  mantissa,  subtract  the  positive  part  of 
the  characteristic  from  9,  giving  the  number  of  ciphers  to  be  placed  between 
the  decimal  point  and  first  significant  figure. 

3.   Find  the  number  whose  logarithm  is  8.5265  — 10. 

5265 
Next  less  mant.  =  5263 ;  figures  corresponding,  336, 
.     0,^  Tab.  diff.  13)2.00 (.15  =  .2,  nearly. 

.      I.-  L3_ 

•      r.  -  CJ^a^.  70 


410       ADVANCED  COURSE  IN  ALGEBRA 

By  the  above  rule,  there  will  be  one  cipher  to  be  placed  be- 
tween the  decimal  point  and  first  significant  figure. 
The  result  is  .03362. 

The  correction  can  usually  be  depended  upon  to  only  one  decimal 
place ;  the  division  should  be  carried  to  two  places  to  determine  the  last 
figure  accurately. 


EXERCISE  97      1 
Find  the  logarithm  of  :                                      \ 

1.    70.                  4.    .0337.                   7.    .9617. 

10.    .0064685. 

2.    .59.                  5.    82.95.     .               8.    .0003788. 

11.   4072.6. 

3.    98.4.                6.    253.07.                 9.    7.803. 

12.    .013592. 

Find  the  number  corresponding  to : 

13.    1.7782.                           17.   0.8744. 

21.    1.8077. 

14.    8.4314-10.                   18.    3.3565. 

22.    7.6899-10. 

15.    0.6522.                            19.    6.2998  -  10. 

23.  -9.9108-10. 

16.   9.0128-10.                  20.   8.9646-10. 

24.   2.5524. 

APPLICATIONS 

601.  The  approximate  value  of  a  number  in  which  the 
operations  indicated  involve  only  multiplication,  division,  invo- 
lution, or  evolution  may  be  conveniently  found  by  logarithms. 

The  utility  of  the  process  consists  in  the  fact  that  addition 
takes  the  place  of  multiplication,  subtraction  of  division, 
multiplication  of  involution,  and  division  of  evolution. 

In  computations  with  four-place  logarithms,  the  result  cannot  usually 
be  depended  upon  to  more  than  four  significant  figures. 

1.   Find  the  value  of  .0631  x  7.208  x  .51272. 
By  §  589,  log  (.0631  x  7.208  x  .51272) 

=  log  .0631  + log  7.208  +  log  .51272. 

log   .0631=    8.8000-10 
log   7.208=   0.8578 
log  .51272  =    9.7099  -  10 
Adding,  log  of  result  =  19.3677  -  20 

=   9.3677  - 10  (See  Note  below.) 


LOGARITHMS  411 

i     Number  corresponding  to  9.3677  -  10  =  .2332. 

If  the  sum  is  a  negative  logarithm,  it  should  be  written  in  such  a  form 
that  the  negative  portion  of  the  characteristic  may  be  —  10. 
Thus,  19.3677  -  20  is  written  9.3677  -  10. 

336.8 


2.   Find  the  value  of 


7984 


By  §  590,  log  ^  =  log  336.8  -  log  7984. 

log  336.8  =  12.5273  -  10  (See  Note  below.) 
log  7984  =    3.9022 
Subtracting,     log  of  result  =    8.6251  - 10 
Number  corresponding  =  .04218. 

To  subtract  a  greater  logarithm  from  a  less,  or  a  negative  logarithm 
from  a  positive,  increase  the  characteristic  of  the  minuend  by  10,  writ- 
ing —  10  after  the  mantissa  to  compensate. 

Thus,  to  subtract  3.9022  from  2.5273,  write  the  minuend  in  the  form 
12.5273  -  10  ;  subtracting  3.9022  from  this,  the  result  is  8.6251  -  10. 

3.   Find  the  value  of  (.07396)^ 

By  §  591,  log  (.07396)^  =  5  x  log  .07396. 

log  .07396  =  8.8690  -  10 


44.3450  -  50 
=  4.3450  - 10  =  log  .000002213. 

4.   Find  the  value  of  ^.035063.  ^  W/J^^' 

By  §  592,  log  a/.035063  =  -  log  .035063.         ^    ^_ 

log  .035063  =  8.5449  - 10 

3)28.5449  -  30  (See  Note  below.) 
9.5150  - 10  =  log  .3274. " 

To  divide  a  negative  logarithm,  write  it  in  such  a  form  that  the  nega- 
tive portion  of  the  characteristic  may  be  exactly  divisible  by  the  divisor, 
with  —  10  as  the  quotient. 

Thus,  to  divide  8.5449  —  10  by  3,  we  write  the  logarithm  in  the  form 
28.5449  -  30.     Dividing  this  by  3,  the  quotient  is  9.5150  -  10. 


412  ADVANCED   COUllSE   IN   ALGEBRA 

602.   Arithmetical  Complement. 

The  Arithmetical  Complement  of  the  logarithm  of  a  number, 
or,  briefly,  the  Cologarithm  of  the  number,  is  the  logarithm  of 
the  reciprocal  of  that  number. 

Thus,        colog409  =  log-^=logl-log409. 

log  1  =  10.  -  10     (See  Ex.  2,  §  601.) 

log  409=   2.6117 
0-.  colog409=   7.3883-10. 

Again,     colog  .067  =  log—  =  log  1  —  log  .067. 
.067 

log  1  =  10.         -10 

log  .067=   8.8261-10 

.-.  colog  .067=   1.1739. 

It  follows  from  the  above  that  the  cologarithm  of  a  number 
may  he  found  by  subtracting  its  logarithin  from  10  — 10. 

•  The  cologarithm  may  be  found  by  subtracting  the  last  significant  figure 
of  the  logarithm  from  10  and  each  of  the  others  from  9,  —  10  being 
written  after  the  result  in  the  case  of  a  positive  logarithm.  * 

.51384 


Ex.     Find  the  value  of 


8.708  X  .0946 


log       -51^-^^       =  log^.51384  X  -^  X  -^^ 
^  8.708  X  .0946         ^V  ^-^^^     •^946;' 


=  log  .51384 +  log-i-  + log-   ^ 


8.708         °.0946 
=  log  .51384  +  colog  8.708  +  colog  .0946. 

log  .51384  =  9.7109  - 10 
colog  8.708  =  9.0601 -10 
colog  .0946  =  1.0241 

9.7951 -10  =  log  .6239. 

It  is  evident  from  the  above  example  that,  to  find  the  loga- 
rithm of  a  fraction  whose  terms  are  the  product  of  factors,  we 
add  together  the  logarithms  of  the  factors  of  the  numerator,  ayid 
the  cologarithms  of  the  factors  of  the  denomiyiator. 


y 


LOGARITHMS"  ."^  ^    'L   ^         418 


The  value  of  the  fraction  may  be  found  without  using  cologarithms  by .  ^   >  ^ 
the  following  formula  :  '^  I  (,  *;  f 

log -^^^^^        =  log  .51384  -  (log  8.708  +  log  .0946)  (§8  589,  590). 

^8.708  X  .0946         '^  ^    ='  ^  J  vs^        ^        J 

The  advantage  in  using  cologarithms  is  that  the  written  wor]j: ^^q^-»V  7  /  - 
putation  is  exhibited  in  a  more  compact  form.  \  /  kj'^H   ' 


603.   Examples. 

1.   Find  the  value  of  ?^. 
3f 


\^ 


i^#^ 


log  ?^  =  log  2  +  log  -v/S  +  colog  3f     (§  602) 

=  log2  +  ilog5  +  ?colog3.  o,«33^ 

log  2=   .3010 
log  5=   .6990;  ^3=    .2330 

colog  3  =  9.5229 -10;    x  ^  =  9.6024-10 

.1364  =  log  1.369. 

A  negative  number  has  no  common  logarithm  (§  580). 

If  such  numbers  occur  in  computation,  they  may  be  treated 
as  if  they  were  positive,  and  the  sign  of  the  result  determined 
irrespective  of  the  logarithmic  work. 

Thus,  to  find  the  value  of  721.3  x  (-  3.0528),  we  find  the  value  of 
721.3  X  3.0528,  and  put  a  negative  sign  before  the  result. 


2.   rind  the  value  of    *''  ~  '^^^^^ 


"^  \  7.962      3    ^  7.9( 


7.962 

03296 
962 


=  i  (log  .03296  -  log  7.962). 

o 

log  .03296  =  8.5180 -10 

log   7.962  =  0.9010  , 

3)27^6170-30 

9.2057 -10  =  log  .1606. 
The  result  is  -  .1606. 


4:U 


ADVANCED  COURSE  IN  ALGEBRA 


1. 
2. 

5. 

6. 
11. 
13. 
14. 
15. 


EXERCISE  98 

Find  by  logarithms  the  values  of  the  following  : 

3.    (  -  54.375)  X  (  -  .00061488). 


18. 
31. 


33. 

34. 
35. 


7. 


2414.7  X  .09348. 
832.4  X  4.1639. 

51.29 
6.348' 
834.32 

2192.4*  °'    - 

( -.009213)  x(- 73.36) 

(-.0832)  X  2.8087 


.004497 
.09769  ' 
3.3629 
-.75438' 


4.    .38142  X  (-.0053909). 

718  X  (-.02415) 
(-.5157)  X  1420.6* 
.87028  X  3.74 


12. 


(2.514)5. 
(-83.28)3. 

(.035127)4. 


19.   VAm. 


20.    V-  .037368. 


21. 


16.    100*. 


^/7 

Va 


i 


17.    (- .007795) 't^.      23.   -^-IM§. 


V1553. 

(837.5  X  .0094325) ' 

V3929  X  v^6K48 
y/T2TM 


31 

24.  (m]i 

V6937y 


V.05287 


V.374  X  V.0078359 


(-  .0001916)' 


^68i: 


.27556 
3801.4 


38. 


39. 


40. 


9. 


10. 


.0006589  x(- 42.318) 
3.8961  X  .6945  x  .01382 


25 


27. 


4694  X  .00457 

v^.  0009657 

\/.'0049784 

-(.25693)^; 

(-  .8346)^ 


28. 


30. 


/76.1  X  .0593\| 
V       1.307       j 

^r  7.544 


31.4  X  .415 


V5106.5  X  .00003109. 
36.    .83184  X  (.2682)3 X  (56.1)i 
37. 


.0005616  X  V424.65 

(6.73)4  X  (.03194)* 

485.7  X  (.7301)7  xv^ 


1000 


(9. 1273)6  X  (.7095)^ 
3 /r(- .95048)5  x(8473)h 
>L  (- 2080.9)  xvC05^  J 
v/-. 0030 12  X  1955 


V. 04142  x(-.947?^) 

41.^ 


(-.843)8 X  ^17959 x(- 560. fi^?" 
04813  X  (5.6074)^ 
.002988)^ 


(10.115)5  x( 


^x  (.65034)0 
S)^  X  \/73r27J 


LOGARITIIxMS  415 

EXPONENTIAL  AND  LOGARITHMIC  EQUATIONS 

604.  An  Exponential  Equation  is  an  equation  in  which,  the 
unknown  number  occurs  as  an  exponent. 

To  solve  an  equation  of  this  form,  take  the  logarithms  of  both 
members ;  the  result  will  be  an  equation  which  can  be  solved 
by  ordinary  algebraic  methods. 

1.  Given  31"  =  23  ;  find  the  value  of  x. 
Taking  the  logarithms  of  both  members, 

log  (31")  =  log  23 ;  or  x  log  31  =  log  23  (§  591). 

^,  log  23     1.3617      Q-,on_u 

Then,  x  =  ^  &       =  =.9130+. 

'  log  31     1.4914 

2.  Solve  the  equation  .2"  =  3. 

Taking  the  logarithms  of  both  members,  x  log  .2  =  log  3. 

Then,  0.=  !^^=— i^Zi— =  ^iI^  =  -.6825+. 
log  .2      9.3010-10      -.699 

An  equation  of  the  form  a'  =  b  may  be  solved  by  inspection 
if  b  can  be  expressed  as  an  exact  power  of  a. 

3.  Solve  the  equation  16"  =  128. 

We  may  write  the  equation  (2*)"  =  2^  or  2*"  =  21 

•   .       ■  7 

Then,  by  inspection,  4  a?  =  7  ;  and  ^'  =  t* 

If  the  equation  were  16^  = ,  we  could  write  it  (2*)*  =  —  =  2-^ ;  then 

7 

4  X  would  equal  —  7,  and  x  = 

4 

Certain  logarithmic  equations  are  readily  solved  by  aid  of 

the  principles  of  §§  589  to  591. 

4.  Given  2  log^  x  =  m ;  find  the  value  of  x. 

By  §  591,  2  log«  x  =  log«  (^x^)  ;  whence,  log«  (a^)  =  m. 
Then  by  §  580,  a'^  =  x^-  whence,  x=±  Va"*  =±a^. 

5.  Given  log  (ic  +  4)  —  log  x  =  3;  find  the  value  of  x. 
By  §  590,  log  (x  +  4)  -  log  x  =  log  ^^^' 


416  ADVANCED   COURSE  IN   ALGEBRA 

Then,  log^+i  =  3;  and  by  §  581,  W  =  ?i±i. 

X  X 

Therefore,  1000  x  =  x-\-4:,  and  x 


999 


Solve  the  following : 

EXERCISE  99 

1.    13- =  8. 

7.    .2-+5  =  .S---*. 

12. 

fi-V  =  JL, 

2.    .06- =  .9. 

8.    .3-4  =  100. 

\27/       81 

3.    9.347- =  .0625. 

9.    16- =  32. 

13. 

52X-1  =  1.. 
25 

4.  .005038-=  816.3. 

5.  34—1  =  42^+3. 

6.  73-4-2  =.8x. 

10.    32-=    ^  . 
128 

14. 
15. 

16.  Given  a,  r,  and   I ;  derive  the  formula  for  n  (§  529). 

17.  Given  a,  r,  and  /S';  derive  the  formula  for  n. 

18.  Given  a,  Z,  and  S ;  derive  the  formula  for  n. 

19.  Given  r,  I,  and  >S' ;  derive  the  formula  for  n. 

Solve  the  following : 

20.  3  loga  X  =  4  loga  w.  23.  log3  -  log  (x  +  1)  =  - 1. 

21.  log(x-5)-log(2x+l)=2.  24.  logx  +  log  (4  a:  +  3)=  1. 

22.  log  5  +  log  (3  X- 2)  =3.  25.  log2  +  2  logx  =  log  (5  a;  -  2). 

605.   Logarithm  of  a  Number  to  Any  Baseo 

1.  Find  the  logarithm  of  .3  to  the  base  7.    • 
By  §  594, 

^'  logioT  .8451  .8451 

Examples  of  this  kind  may  be  solved  by  inspection,  if  the 
number  can  be  expressed  as  an  exact  power  of  the  base. 

2.  Find  the  logarithm  of  128  to  the  base  16. 
Let  logiel28  =  a;;  then,  by  §  580,  16*  =  128. 

Then,  as  in  Ex.  3,  §  604,  a;  =  - ;  that  is,  logig  128  =  -  • 

4  4 


LOGARITHMS  417 

EXERCISE  100 

Find  the  values  of  the  following  : 

1.  log7  59.  3.    log.482.  5.    log68  2.915. 

2.  log6.7.  4.    log.9 .00453.  6.    logoi  .06038. 
Find  by  inspection  the  values  of  the  following : 

7.    log.5l25.         8.    log,.(l).         9.   log^,^(3).  10.    log^^(^)- 

EXPONENTIAL  AND  LOGARITHMIC  SERIES- 
606.   Let  n  be  a  real  number  greater  than  unity. 

By  §364,  [{^  +  iy]=(i+iy- 

Expanding  both  members  by  the  Binomial  Theorem, 

L  ^  [2  n'  [3  n'^      J 

n  [2  n^ 

nx(nx—  l)(nx  —  2)     J^  .^. 

Since,  by  hypothesis,  n  is  >  1,  -  is  numerically  <  1 ;  and 

both  members  of  (1)  are  convergent  series  (§  559). 
We  may  write  equation  (1)  in  the  form 


[2  [3 


■J....] 


f        1\        f        l\f        2\ 

x[x x[x a; 

which  holds  however  great  n  may  be. 

Now  let  71  be  indefinitely  increased. 

12 
Then,  the  limit  of  each  of  the  terms  -,  -,  etc.,  is  0  (§  248). 

n    n 


418  ADVANCED   COURSE   IN   ALGEBRA 

Hence,  the  limiting  value  of  the  first  member  of  (2)  is 

and  the  limiting  value  of  the  second  member  is 

'  +  -'"  +  [2  +  g  +  -- 

By  §  252,  these  limits  are  equal ;  that  is, 

x^  .  X 


}+'+tt 


The  series  in  the  second  member  is  convergent  for  every 
value  of  X  (§  558)  ;  and  the  series  in  brackets  is  also  conver- 
gent, for  it  is  obtained  from  the  series  in  the  second  member 
by  putting  1  in  place  of  ic.  . 

Denoting  the  series  in  brackets  by  e,  we  have 

which  holds  for  every  value  of  x. 
Putting  mx  for  x,  in  (3), 

e"'^  =  l  +  ma;+— —  +  -—-  +  ....  (4) 

\A         Lr 

Let  m  =  logg  a,  where  a  is  any  positive  real  number. 
Then  e'"  =  a  (§  580),  and  e'^'  =  a\ 
Substituting  these  values  in  (4),  we  obtain 

a'  =  l+(log.a)x  +  (log.a)^|  +  (log.a)'g+...;  (5) 

which  holds  for  all  values  of  x,  and  all  positive  real  values 
of  a. 

The  result  (5)  is  called  the  Exponential  Series. 

607.  The  system  of  logarithms  which  has  e  for  its  base 
is  called  the  Napierian  System,  from  Napier,  the  inventor 
of  logarithms. 

Napierian  logarithms  are  also  called  Natural  Logarithms. 


LOGARITHMS  419 

The  approximate  value  of  e  may  be  readily  calculated  from 
the  series  of  §  606, 

and  will  be  found  to  equal  2.7182818.... 

608.    To  expand  log^  (1  +  x)  in  ascending  powers  of  x. 

Substituting  in  (5),  §  606,  l-\-x  for  a,  and  y  for  x, 

(1  +  a;)^  =  1  4-  [loge  (1  +  ic)]  ?/  4-  terms  in  y^,  /,  etc. ; 
which  holds  for  all  values  of  y,  provided  x  is  real,  and  alge- 
braically greater  than  —  1. 

Expanding  the  first  member  by  the  Binomial  Theorem, 

=  1  +  [log,  (1  +  i»)]  2/  +  terms  in  y%  y%  etc.  (6) 

The  first  membet  of  (6)  is  convergent  when  x  is  numerically 
less  than  1  (§  559). 

Hence,  (6)  holds  for  all  values  of  y,  provided  x  is  real,  and 
numerically  less  than  1. 

Then,  by  the  Theorem  of  Undetermined  Coefficients,  the 
coefficients  of  y  in  the  two  series  are  equal ;  that  is, 

l^        l£        11 
Or,  log,(l  +  a^)  =  a^-f  +  f-f  +  f--;  (7) 

which  holds  for  all  values  of  x  numerically  less  than  1. 
This  result  is  called  the  Logarithmic  Series. 

It  was  proved  in  §  557  that  this  series  was  convergent  when  x  was 
numerically  less  than  1. 

It  was  also  shown  in  §  556  that  it  was  convergent  when  x  =  l,  and  in 
§  548  that  it  was  divergent  when  x  =  —  1. 

Then,  series  (7)  can  be  used  to  calculate  Napierian  logarithms,  pro- 
vided X  is  taken  equal  to,  or  numerically  less  than,  1. 

Unless  X  is  small,  it  requires  the  sum  of  a  great  many  terms  to  ensure 
any  degree  of  accuracy. 


420 


ADVANCED   COURSE  IN  ALGEBRA 


609.   We  will  now  derive  a  more  convenient  series  for  the 
calculation  of  Napierian  logarithms. 

Putting  —  X  for  x,  in  (7),  §  608,  we  have 

^  2      3      4      5 

Subtracting  (8)  from  (7),  we  obtain 

»log,(l  +  a.')-log,(l-a^)  =  2a^  +  ?^  +  ?^+.... 


(8) 


Or  (§  590), 


lo^^l±l  =  2(x  +  t  +  t 
^"1-x        \        3      5 


(9) 


Let  X  = ;  m  and  n  being  positive,  and  m  >  n. 


m-\-n' 


m  —  n 


This  IS  a  valid  substitution,  since  in  this  case  — — -  <  1. 

m-{-n 

m  —  71 


Then, 


1  + 
1  —  x     -i      m  —  n      2  7i~  n 


Substituting  these  values  in  (9),  we  obtain 

2      m  _  Srti  —  n^      \fm  —  nV     l/m  —  nV 
"n         _m-{-n     6\m-\-nj       5\m-{-nJ 

But  by  §  590,  log^—  =  log^  m  —  log«  n ;  whence, 


loge  m  =  loge  n  +  2 


"m  —  ^*  I  1/wi  —  n\^  .  1/  m 


m  +  n     S\m  +  ny      5V  m  + 


^)'-} 


610.     Let   it  be   required,   for   example,   to   calculate    the 
Napierian  logarithm  of  2  to  six  places  of  decimals. 
Putting  m  =  2  and  w  =  1  in  the  result  of  §  609,  we  have 


log,2  =  log,l  +  2 


hm-w^-] 


Or  since  log,l  =  0  (§585), 

.  log,  2  =  2(.3333333  +  .0123457  +  .0008230  +  .0000653 
+  .0000056  -f  .0000005  +  •••) 
=  2  X. 3465734  =  .6931468. 


LOGARITHMS  421 

Then,  log^  2  =  .693147,  to  the  nearest  sixth  place  of  decimals. 
Having  found  log^2,  we  may  calculate  log^3  by  putting  m=3 
and  w  =  2  in  the  result  of  §  609. 

Proceeding  in  this  way,  we  shall  find  log^  10  =  2.302585 

611.  To  calculate  the  common  logarithm  of  a  number,  having 
given  its  Napierian  logarithm. 

Putting  6  =  10  and  a  =  e  in  the  result  of  §  594, 

^^^^«"  =  iS  ==  2:3485  ^  log.m  =  .4342945  x  log.m. 
Thus,         logio2  =  .4342945  x  .693147  =  .301030. 

612.  The  multiplier  by  which  logarithms  of  any  system  are 
derived  from  Napierian  logarithms,  is  called  the  modulus  of 
that  system. 

Thus,  .4342945  is  the  modulus  of  the  common  system. 

Conversely,  to  find  the  Napierian  logarithm  of  a  number 
when  its  common  logarithm  is  given,  we  may  either  divide  the 
common  logarithm  by  the  modulus  .4342945,  or  multiply  it  by 
2.302585,  the  reciprocal  of  .4342945. 

EXERCISE    101 

Using  the  table  of  common  logarithms,  find  the  Napierian  logarithm  of 
each  of  the  following  to  four  significant  figures  : 

1.  1000.  3.  9.93.  6.    .04568. 

2.  .0001.  4.   243.6.  6.   .56734. 

7.  What  is  the  characteristic  of  logs  758  ? 

8.  What  is  the  characteristic  of  logy  500  ? 

9.  If  log  3  =  .4771,  how  many  digits  are  there  in  31^  ? 

10.  If  log  8  =  .9031,  how  many  digits  are  there  in  828  ? 

11.  If  log  11  =  1.0414,  how  many  digits  are  there  in  the  integral 
part  of  11"^? 


422       ADVANCED  COURSE  IN  ALGEBRA 


XXX.   COMPOUND  INTEREST  AND 
ANNUITIES 

613.  The  principles  of  logarithms  may  be  applied  to  the 
solution  of  problems  in  Compound  Interest. 

Let  P  =  number  of  dollars  in  the  principal ; 

n  =  number  of  years ; 

t  =  the  ratio  to  one  year  of  the  time  during  which  sim- 
ple  interest   is  calculated;   thus,  if   interest   is 
compounded  semi-annually,  ^  =  ^  5 
M  =  number  of  dollars  in  the  amount  of  one  dollar  for 
t  years ; 

A  =  number  of  dollars  in  the  amount  of  P  dollars  for 
n  years. 

Since  one  dollar  amounts  to  JR  dollars  in  t  years,  P  dollars 
will  amount  to  PE  dollars  in  t  years ;  that  is,  the  amount  at 
the  end  of  the  1st  interval  is  PR  dollars. 

In  like  manner,  the  amount  at  the  end  of  the 
2d  interval  is  PE  x  E,  or  PE'^  dollars ; 
3d  interval  is  PE^  X  E,  or  PE^  dollars  ;  etc. 


Since  the  whole  number  of  intervals  is  -,  the  amount  at 
the  end  of  the  last  one,  in  accordance  with,  the  law  observed 
above,  will  be  PE^  dollars. 

n 

That  is,  A  =  PEl  (1) 

By  §  §  589,  591,         log  ^  =  log  P  +  'i  log  E.  (2) 

1.  What  will  be  the  amount  of  $7326  for  3  years  and  9 
months  at  7  per  cent  compound  interest,  interest  being  com- 
pounded quarterly  ? 

Here,  P  =  7326,  n  =  3|,  i  =  ^  i?  =  1.0175,  ^  =  15. 


COMPOUND  INTEREST   AND  ANNUITIES  423 

log  P  =  3.8649 
log  i2  =  0.0075;    X  15  =  0.1125 

log  ^  =  3.9774    .-.^  =  $9493. 

2.  What  sum  of  money  will  amount  to  ^  1763.50  in  3  years 
at  5  per  cent  compound  interest,  interest  being  compounded 
semi-annually  ? 

From  (2),  log  P  =  log  ^  -  -  log  R. 

Here,  n  =  ^,t  =  \,R  =  1.025,  A  =  1763.5,  -  =  6. 

Z  t 

log  A  =  3.2464 
log  J?  =  0.0107;  X  6  =  0.0642 

.  log  P  -  3.1822     .-.  P  =  $  1521. 

3.  In  how  many  years  will  ^300  amount  to  $398.60  at  6 
per  cent  compound  interest,  interest  being  compounded  quar- 
terly ? 

From  (2),  ^^t(log  A-log  P)^ 

log  B 

Here,  P=  300,  t  =  j,  P  =  1.015,  ^  =  398.6. 

^  log  398.6  -  log  300  ^  2.6006  -  2.4771  ^  .1235 
***  *^  ~        4  log  1.015  4  X  .0065  .0260 

=  4.75  years. 

4.  At  what  rate  per  cent  per  annum  will  $  500  amount  to 
^.83  in  6  years  and  6  months,  interest  being  compounded 

semi-annually  ? 


^C^)> 

\logi?  = 

_  — o  "        --0 

n 

,p= 

500, 

n  =  6i,  t  = 

t 
=  1  ^  =  688.83, 

n 
1 

=  13. 

log^  = 
logP  = 

log  12  = 

2.8381 

2.6990 

13)0.1391 

0.0107     .-.   R 

1.025. 

424  ADVANCED   COURSE  IN  ALGEBRA 

That  is,  the  interest  on  one  dollar  for  6  months  is  $  .025,  and 
the  rate  is  5  per  cent  per  annum. 

614.  To  find  the  present  worth  of  A  dollars  due  at  the  end 
of  n  years,  interest  being  comjjounded  annually. 

Putting  ^  =  1  in  (1),  §  613,  we  have 

A  =  PBT;  whence,  ^  =  ^• 

ANNUITIES 

615.  An  Annuity  is  a  fixed  sum  of  money  payable  at  equal 
intervals  of  time. 

In  the  present  chapter  we  shall  consider  those  cases  only  in 
which  the  payments  are  annual ;  in  finding  the  present  worth 
of  such  an  annuity,  it  is  customary  to  compound  interest  annu- 
ally. 

When  we  speak  of  the  annuity  as  beginning  at  a  certain 
epoch,  it  is  understood  that  the  first  payment  becomes  due 
one  year  from  that  time. 

616.  To  find  the  present  worth  of  an  annuity  to  continue  for 

n  successive  years,  allowing  compound  interest. 

Let  A   =  number  of  dollars  in  the  annuity ; 

E  =  number  of  dollars  in  the  amount  of  one  dollar  for 
one  year ; 

P  =  number  of  dollars  in  the   present  worth  of  the 
annuity. 

By  §  614,  the  present  worth  of  the 

1st  payment  =  —; 
H 

2d  payments—; 


nth.  payment  =  —  • 


I 


COMPOUND  INTEREST   AND  ANNUITIES  425 


Hence,  the  sum  of  the  present  worths  of  the  separate  pay- 
ments, or  the  present  worth  of  the  annuity,  is 


E^     E''-'  E'     E 


That  is,      P=A 


R-  ^  i2«-i  ^     ^  E'     EJ 


The  expression  in  brackets  is  the  sum  of  the  terms  of  a 

geometric  progression,  in  which  a  =  —-,  r  =  E,  and  I  =  — • 

it"  E 

Then,  by  II,  §  527,       P=^-^_^.  (3) 

Ex.     Find  the. present  worth  of  an  annuity  of  $150  to  con- 
tinue for  20  years,  allowing  4  per  cent  compound  interest. 

Here,  A  =  150,  n  =  20,  E  =  1.04,  E-l  =  .04. 


Whence,  ^=^'[l "  (di).} 

l°g^r^  =  20cologl.04. 

colog  1.04  =  9.9830 -10 

20 

9.6600  -  10  =  log  .4571. 

Then,  P=  ^(1  -  .4571)  =  3750  x  .5429. 

log  3750  =  3.5740 
log  .5429  =  9.7347 -10 

log  P=  3.3087  .-.  P=$2036. 

617.   We  have  from  (3),  §  616, 

^_P(Ii-l)_PE^(E-l), 
^-    ^_\_    -      E^-1      '  (4) 

E- 

which  is  a  formula  for  finding  the  annuity  to  continue  for  n 
successive  years,  when  the  present  worth  and  the  amount  of 
one  dollar  for  one  year  are  given. 


426  ADVANCED  COURSE  IN  ALGEBRA 

Formula  (4)  may  also  be  used  to  find  what  fixed  annual  payment  must 
be  made  to  cancel  a  note  of  P  dollars  due  n  years  hence,  B  being  the 
number  of  dollars  in  the  amount  of  one  dollar  for  one  year. 

618.  If  in  (3),  §  616,  n  be  indefinitely  increased,  the  limit- 
ing value  of  the  second  member  is 

A 


R-1 


(§  248). 


That  is,  the  present  worth  of  a  perpetual  annuity  is  equal  to 
the  amount  of  the  annuity  divided  by  the  interest  on  one  dollar 
for  one  year. 

619.  To  find  the  present  worth  of  an  annuity  to  begin  after 
m  years,  and  continue  for  n  years,  allowing  compound  interest. 

With  the  notation  of  §  616,  the  number  of  dollars  in  the 
value  of  the  annuity  one  year  before  the  first  payment  becomes 
due,  is  ^ 

2Z^    or   4(^!^li}. 
By  §  614,  the  present  worth  of  this  amount,  due  m  years 

R\R-1)  ' 

Therefore,  P  =  ^^^^/"^^  » 

'  R'^+XR  -  1) 

620.  By  §  618,  the  present  worth  of  a  perpetual  annuity  to 
begin  after  m  years,  is  given  by  the  formula 

A 


P  = 


R^{R  - 1) 


EXERCISE  102 

1.  What  will  be  the  amount  of  $1300  for  16  years  at  5  per  cent 
compound  interest,  the  interest  being  compounded  annually  ? 

2.  What  sum  of  money  will  amount  to  $981.75  in  8  years  and  9 
months  at  4  per  cent  compound  interest,  the  interest  being  compounded 
quarterly  ? 


COMPOUND   INTEREST  AND  ANNUITIES         427 

3.  In  how  many  years  will  $859  amount  to  $1012.80  at  3  per  cent 
compound  interest,  the  interest  being  compounded  semi-annually  ? 

4.  What  is  the  present  worth  of  a  note  for  $625.34  due  12  years 
hence,  allowing  3|  per  cent  compound  interest,  the  interest  being  com- 
pounded annually  ? 

6.  At  what  rate  per  cent  per  annum  will  $3700  gain  $678  in  4  years 
and  3  months,  the  interest  being  compounded  quarterly  ? 

6.  In  how  many  years  will  a  sum  of  money  double  itself  at  6  per 
cent  compound  interest,  the  interest  being  compounded  annually  ? 

7.  In  how  many  years  will  a  sum  of  money  treble  itself  at  4|  per 
cent  compound  interest,  the  interest  being  compounded  semi-annually  ? 

8.  What  is  the  present  worth  of  an  annuity  of  $  300  to  continue  14 
years,  allowing  4  per  cent  compound  interest  ? 

9.  What  is  the  present  worth  of  a  perpetual  annuity  of  $506.70,  allow- 
ing 3^  per  cent  compound  interest  ? 

10.  What  is  the  present  worth  of  an  annuity  of  $2238  to  continue  4 
years,  allowing  6  per  cent  compound  interest  ? 

11.  What  is  the  present  worth  of  an  annuity  of  $2680  to  begin  after 
10  years  and  continue  for  7  years,  allowing  5  per  cent  compound  interest  ? 

12.  What  fixed  annual  payment  must  be  made  in  order  to  cancel  a 
note  for  $3500  in  5  years,  allowing  4|  per  cent  compound  interest? 

13.  What  is  the  present  worth  of  a  perpetual  annuity  of  $297.50,  to 
begin  after  8  years,  allowing  3|  per  cent  compound  interest  ? 

14.  What  annuity  to  continue  12  years  can  be  purchased  for  $3149, 
allowing  7  per  cent  compound  interest  ? 

15.  A  person  borrows  $6365  ;  how  much  must  he  pay  in  annual 
instalments  in  order  that  the  whole  debt  may  be  discharged  in  10  years, 
allowing  4^  per  cent  compound  interest  ? 


428  ADVANCED  COURSE  IN  ALGEBRA 


XXXI.    PERMUTATIONS  AND  COMBINA- 
TIONS 

621.  The  different  orders  in  which  things  can  be  arranged 
are  called  their  Permutations. 

Thus,  the  permutations  of  the  letters  a,  b,  c,  taken  two  at  a 
time,  are  ab,  ac,  bal  be,  ca,  cb ;  and  their  permutations  taken 
three  at  a  time,  are  abc,  acb,  bac,  bca,  cab,  cba. 

622.  The  Combinations  of  things  are  the  different  collections 
which  can  be  formed  from  them  without  regard  to  the  order 
in  which  they  are  placed. 

Thus,  the  combinations  of  the  letters  a,  b,  c,  taken  two  at  a 
time  are  ab,  be,  ca;  for  though  ab  and  ba  are  different  permu- 
tations, they  form  the  same  combination. 

623.  To  find  the  number  of  permutations  of  n  different  things 
taken  two  at  a  time. 

Consider  the  n  letters  a,  b,  c,  •••. 

In  making  any  particular  permutation  of  two  letters,  the 
first  letter  may  be  any  one  of  the  n ;  that  is,  the  first  place 
can  be  filled  in  n  different  ways. 

After  the  first  place  has  been  filled,  the  second  place  can  be 
filled  with  any  one  of  the  remaining  7i  —  1  letters. 

Then,  the  whole  number  of  permutations  of  the  letters  taken 
two  at  a  time  is  n(7i  —  l). 

We  will  now  consider  the  general  case. 

624.  To  find  the  number  of  permutations  ofn  different  things 
taken  r  at  a  time. 

Consider  the  n  letters  a,  b,  c,  -". 

In  making  any  particular  permutation  of  r  letters,  the  first 
letter  may  be  any  one  of  the  n. 

After  the  first  place  has  been  filled,  the  second  place  can  be 
filled  with  any  one  of  the  remaining  n  —  1  letters. 


PERMUTATIONS   AND   COMBINATIONS  429 

After  the  second  place  has  been  filled,  the  third  place  can  be 
filled  in  n  —  2  different  ways. 

Continuing  in  this  way,  the  rth  place  can  be  filled  in 

n  —  (r  —  1),  or  n  —  r-\-l  different  ways. 

Then,  the  whole  number  of  permutations  of  the  letters  taken 
r  at  a  time  is  given  by  the  formula 

„P,  =  7i(n  _  1)  (n -  2)  ... (ii - r  + 1).  (1) 

The  number  of  permutations  of  n  different  things  taken  r  at  a  time  is 
usually  denoted  by  the  symbol  nPr- 

625.  If  all  the  letters  are  taken  together,  r  =  n,  and  (1) 

becomes  ^P^  =  n{n-l)(n-2) -3 -2 'l  =  \n.  (2) 

Hence,  the  number  of  permutations  of  n  different  things  taken 
n  at  a  time  equals  the  product  of  the  natural  numbers  from  1  to  n 
inclusive. 

626.  To  find  the  number  of  combinations  of  n  different  things 
takeri  r  at  a  time.        ^ 

The  number  of  permutations  of  n  different  things  taken  r  at 
a  time,  is      ^^  (^  - 1)  (n  -  2) ...  (71  -  r  + 1)  (§  624).      .^    , 

But  by  §  625,  each  combination  of  r  different  things  may 
have  I  r  permutations.     '^'^"'^'^^J^  ^ 

Hence,  the  number  of  combinations  of  n  different  things 
taken  r  at  a  time  equals  the  number  of  permutations  divided 
by  \r.  p 

That  is,       ^o^^^(^-l)(^-2)-(n-r  +  l).    ^^1^^^) 

The  number  of  combinations  of  n  different  things  taken  r  at  a  tinit^  is 
usually  denoted  by  the  symbol  nCr- 

627.  Multiplying  both  terms  of  the  fraction  (3)  by  the  prod- 
uct of  the  natural  numbers  from  1  to  71  —  r  inclusive,  we  have 

^  ^n{n-V)'"(n-r^l)»{n-r)-"2'l^      J!^       . 
\rxl'2"-(n-r)  \r  \n-r' 

which  is  another  form  of  the  result. 


430  ADVANCED  COUKSE  IN   ALGEBRA 

628.  Tlie  number  of  combinations  of  n  different  things  taken  r 
at  a  time  equals  the  number  of  combinations  taken  n  —  r  at  a  time. 

For  in  making  a  selection  of  r  things  out  of  ri,  we  leave  a 
selection  of  n  —  r  things. 

The  theorem  may  also  be  proved  by  using  the  result  of  §  627. 

629.  Examples. 

1.  How  many  changes  can  be  rung  with  10  bells,  taking  7 
at  a  time  ? 

Putting    n  =  10,  r  =  7,  in  (1),  §  624, 

ioPy.=  10  .  9  .  8  .  7  .  6  .  5  .  4  =  604800. 

2.  How  many  different  combinations  can  be  formed  with  16 
letters,  taking  12  at  a  time  ? 

By  §  628,  the  number  of  combinations  of  16  different  things, 
taken  12  at  a  time,  equals  the  number  of  combinations  of  16 
different  things,  taken  4  at  a  time. 

Putting  n  ==  16,  r=4,  in  (3),  §  626, 

p      16  .  15  ♦  14  .  13     .^^^ 
''^'^     1.2.3.4     =^^^^- 

3.  How  many  different  words,  each  consisting  of  4  consonants 
and  2  vowels,  can  be  formed  from  8  consonants  and  4  vowels  ? 

The  number  of  combinations  of  the  8  consonants,  taken  4  at 

a  time,  is  ^^1^1^,  or  70. 

The  number  of  combinations  of  the  4  vowels,  taken  2  at  a 
time,  is  q-^,  or  6. 

Any  one  of  the  70  sets  of  consonants  may  be  associated  with 
any  one  of  the  6  sets  of  vowels ;  hence,  there  are  in  all  70  x  6, 
or  420  sets,  each  containing  4  consonants  and  2  vowels. 

But  each  set  of  6  letters  may  have  |6,  or  720  different 
permutations  (§  625). 

Therefore,  the  whole  number  of  different  words  is 

420  X  720,  or  302400. 


<^ 


PERMUTATIONS   AND   COMBINATIONS  431 

EXERCISE  103 

Find  the  values  of  the  followhig  : 

1.  14^6.  3.  17  P7.  5.   17  On. 

2.  9P9.  4.  \5G^.  ^  6.  29C'24. 

7.  In  a  certain  play,  there  are  five  parts  to  be  taken  by  a  company  of 
twelve  persons.     In  how  many  different  ways  can  they  be  assigned  ? 

8.  How  many  different  words,  of  nine  different  letters  each,  can  be 
formed  from  the  letters  in  the  word  flowering^  if  the  vowels  retain  their 
places  ? 

9.  How  many  different  numbers,  of  seven  different  figures  each,  can  /^ 
be  formed  from  the  digits  1,  2,  3,  4,  5,  6,  7,  8,  9,  if  e5ch  number  begins  with  ^  tT 

I  and  ends  with  9  ? 

10.  How  many  even  numbers,  of  five  different  figures  each,  can  be 
formed  from  the  digits  4,  5,  6,  7,  8  ?  ^ 

11.  How  many  different  committees,  of  8  persons  each,  can  be  formed 
from  a  corporation  of  14  persons  ?      In  how  many  will  any  particular  ■- — __^ 
individual  be  found  ?     In  how  many  will  any  particular  individual  be 
excluded  ? 

12.  A  and  B  are  in  a  company  of  72  men.  If  the  company  is  divided 
into  squads  of  6,  in  how  many  of  them  will  A  and  B  be  in  the  same  squad  ? 

13.  In  how  many  different  ways  can  six  persons  be  seated  at  a  round 
table  ? 

14.  There  are  15  points  in  a  plane,  no  three  in  the  same  straight  line. 
How  many  quadrilaterals  can  be  formed,  having  four  of  the  points  for 
vertices  ? 

15.  If  32  soldiers  are  drawn  up  in  line  4  deep,  in  how  many  different 
ways  can  they  be  arranged  so  as  to  have  a  different  set  in  the  front  rank  ? 
In  how  many  ways,  if  the  front  rank  is  always  to  contain  3  particular 
men? 

16.  If  the  number  of  combinations  of  2  n  different  things  taken  w  — 1  at 
a  time,  is  to  the  number  of  combinations  of  2/i  — 2  different  things  taken 

II  at  a  time  as  132  :  35,  find  the  value  of  n. 

17.  How  many  different  crews,  each  consisting  of  eight  oarsmen  and  a 
steersman,  can  be  formed  from  16  boys,  of  whom  12  can  row  but  cannot 
steer,  and  the  others  can  steer  but  cannot  row  ? 

18.  A  person  has  22  acquaintances,  of  whom  14  are  males.  In  how 
many  ways  can  he  invite  17  guests  from  them  so  that  10  may  be  males  ? 

19.  Out  of  10  soldiers  and  15  sailors,  how  many  different  parties  can  be 
formed,  each  consisting  of  3  soldiers  and  3  sailors  ? 


432  ADVANCED  COURSE  IN   ALGEBRA 

20.  From  3  sergeants,  8  corporals,  and  16  privates,  how  many  different 
parties  can  be  formed,  each  consisting  of  1  sergeant,  2  corporals,  and  5 
privates  ? 

21.  Out  of  3  capitals,  6  consonants,  and  4  vowels,  how  many  different 
words  of  six  letters  each  can  be  formed,  each  beginning  with  a  capital, 
and  having  3  consonants  and  2  vowels  ? 

22.  How  many  different  words  of  8  letters  each  can  be  formed  from 
eight  letters,  if  4  of  the  letters  cannot  be  separated  ?  How  many  if 
these  four  can  only  be  in  one  order  ? 

23.  In  how  many  different  ways  can  ten  soldiers  be  drawn  up  in  double 
rank,  if  three  particular  men  are.  always  in  the  front  rank,  and  three  others 
always  in  the  rear  ? 

24.  How  many  different  numbers  of  seven  figures  each  can  be  formed 
from  the  digits  1,  2,  3,  4,  5,  6,  7,  8,  9,  if  the  first,  fourth,  and  last  digits 
are  odd  numbers  ? 

25.  How  many  different  words  of  six  letters  each  can  be  formed  from 
the  letters  in  the  woid  percolating,  if  each  word  has  a  consonant  for  its 
first  and  last  letter,  and  a  vowel  for  its  second  and  fifth  ? 

26.  There  are  2  n  guests  at  a  dinner-party.  If  the  host  and  hostess  have 
fixed  places  opposite  to  each  other,  and  two  specified  guests  cannot  sit  next 
each  other,  in  how  many  ways  can  the  company  be  seated  ? 

630.  To  find  the  number  of  permutations  of  n  things  which  are 
not  all  different,  taken  all  together. 

Let  there  be  n  letters,  of  which  p  are  a's,  q  are  6's,  and  r  are 
c's,  the  rest  being  all  different. 

Let  JSf  denote  the  number  of  permutations  of  these  letters 
taken  all  together. 

Suppose  that,  in  any  assigned  permutation  of  the  n  letters, 
the  p  a's  were  replaced  by  p  new  letters,  differing  from  each 
other  and  also  from  the  remaining  n~p  letters. 

Then  by  simply  altering  the  order  of  these  p  letters  among 
themselves,  without  changing  the  positions  of  any  of  the  other 
letters,  we  could  from  the  original  permutation  form  [p  differ- 
ent permutations  (§  625). 

If  this  were  done  in  the  case  of  each  of  the  N  original  per- 
mutations, the  whole  number  of  permutations  would  be 

Nx[p. 


PERMUTATIONS   AND  COMBINATIONS  433 

Again,  if  in  any  one  of  the  latter  the  q  6's  were  replaced  by 
q  new  letters,  differing  from  each  other  and  from  the  remaining 
n  —  q  letters,  then  by  altering  the  order  of  these  q  letters  among 
themselves,  we  could  from  the  original  permutation  form  \q 
different  permutations ;  and  if  this  were  done  in  the  case  of 
each  of  the  N x\p  original  permutations,  the  whole  number  of 
permutations  would  be  ^  x  Lp  X  |^. 

In  like  manner,  if  in  each  of  the  latter  the  r  c's  were  re- 
placed by  r  new  letters,  differing  from  each,  other  and  from 
the  remaining  n  —  r  letters,  and  these  r  letters  were  permuted 
among  themselves,  the  whole  number  of  permutations  would  be 
Nx\jpx\qx\r. 

But  the  number  of  permutations  on  the  hypothesis  that  the 
n  letters  are  all  different,  is  \n  (§  625). 

\n 
Therefore,  JVx  |i>  X I g  X  |r  =  1^1 :  or,  iV= — =— * 
L_    L_    L_    i_-  Lp[2[>; 

Any  other  case  may  be  treated  in  a  similar  manner. 

Ex.  How  many  permutations  can  be  formed  from  the  letters 
in  the  word  Tennessee,  taken  all  together  ? 

Here  there  are  4  e's,  2  n's,  2  s's,  and  1  t. 

Putting  in  the  above  formula  n  =  9,  p  =  4,  g  =  2,  r  =  2,  we 
have 

\9     _5.6.7.8.9^3,g()^ 


[4[2[2  2-2 

631.  To  find  the  number  of  permutations  ofn  different  things, 
taken  r  at  a  time,  vjhen  each  may  occur  any  number  of  times 
from  once  up  to  r  times,  inclusive. 

Consider  the  n  letters  a,  b,  c,  •••. 

In  making  any  particular  permutation  of  r  letters,  the  first 
letter  may  be  any  one  of  the  ii ;  that  is,  the  first  place  can  be 
filled  in  n  different  ways. 

The  second  letter  can  also  be  any  one  of  the  n ;  that  is,  the 
second  place  can  be  filled  in  n  different  ways. 


434  ADVANCED  COURSE  IN   ALGEBRA 

Continuing  in  this  way,  the  rth  place  can  be  filled  in  71  differ- 
ent ways.  f 

Then,  the  number  of  permutations  is 

71  xnx  "'  to  7^  factors,  or  n"". 

Ex.  How  many  different  words  of  four  letters  each  can  be 
formed  from  nine  letters,  if  each  letter  may  occur  any  number 
of  times  from  once  up  to  four  times,  inclusive  ? 

Here,  n  =  9,  r  =  4. 

Then,  the  number  of  different  words  is 

9^  or  6561. 

632.  To  find  the  entire  number  of  combinatio^is  of  n  different 
things  J  when  each  may  he  taken  any  number  of  times  from  once  to 
n  times,  inclusive. 

Consider  the  n  letters  a,  6,  c,  •••. 

We  can  take  them  one  at  a  time  in  71  ways. 

We  can  take  them  2  at  a  time  in  '^('^-^)  ^^ays  (§  626). 


We  can  take  them  all  in  one  way. 
Then,  the  entire  number  of  ways  is 

n  +  '-^i^^y^  +  -+l  =  2"-l  (§289,1). 

\A 

Ex.   In  how  many  ways  can  a  selection  of  one  or  more  vol- 
umes be  made  from  5  books  ? 

Here,  n  =  5 ;  then,  the  entire  number  of  selections  is 
2^-1,  or  31. 

EXERCISE  104 

1.  In  how  many  different  orders  may  the  letters  of  the  word  denomi- 
nation be  written  ? 

2.  There  are  four  white  billiard  balls  exactly  alike,  and  three  red  balls, 
also  alike.     In  how  many  different  ways  can  they  be  arranged  ? 


PERMUTATIONS  AND  COMBINATIONS  435 

3.  In  how  many  ways  can  six  things  be  given  to  five  persons,  if  there 
is  no  restriction  as  to  the  number  each  may  receive  ? 

4.  How  many  different  numbers  less  than  10000  can  be  formed  from 
the  digits  1,  2,  3,  4,  5,  6,  7,  8  ? 

5.  In  how  many  different  orders  may  the  letters  of  the  word  indepen- 
dence be  written  ? 

6.  How  many  different  signals  can  be  made  with  7  flags,  of  which  2 
are  blue,  3  red,  and  2  white,  if  all  are  hoisted  for  each  signal  ? 

-  7.  A  railway  signal  has  m  arms,  and  each  can  be  placed  in  n  positions. 
How  many  different  signals  can  be  made  with  it  ? 

8.  A  man  has  eight  friends.  In  how  many  ways  can  he  invite  one  or 
more  of  them  to  dinner  ? 

9.  How  many  different  words  of  eight  letters  each  can  be  formed  from 
the  letters  in  the  word  arranged^  if  the  first,  fourth,  and  seventh  letters 
are  always  vowels  ? 

10.  A  house  has  nine  windows  in  front.  How  many  different  signals 
can  be  given  by  having  one  or  more  of  the  windows  open  ? 

11.  In  how  many  ways  can  13  books  be  arranged  on  a  shelf,  when*  five 
volumes  are  alike,  and  four  other  volumes  are  also  alike  ? 

12.  How  many  different  numbers  greater  than  1000000  can  be  formed 
from  the  digits  4,  3,  3,  3,  2,  2,  0  ? 

13.  In  how  many  ways  can  two  dimes,  three  quarters,  four  halves, 
and  five  dollars  be  distributed  among  14  persons,  so  that  each  may  receive 
a  coin  ? 

14.  A  bag  contains  a  cent,  a  half-dime,  a  dime,  a  twenty-cent  piece,  a 
quarter-dollar,  a  half-dollar,  and  a  dollar.  In  how  many  ways  can  a  sum 
of  money  be  drawn  from  the  bag  ? 


436  ADVANCED  COURSE   IN  ALGEBRA 


XXXII.    PROBABILITY 

633.  Suppose  that  a  bag  contains  5  white  balls,  4  red  balls, 
and  3  black  balls,  and  that  one  ball  is  drawn  at  random. 

Any  one  ball  is.  as  likely  to  be  drawn  as  any  other. 

The  drawing  of  a  hall  can  occur  in  12  different  ways ;  for  any 
one  of  the  balls  may  be  drawn. 

The  drawing  of  a  white  hall  can  occur  in  5  different  ways ; 

for  any  one  of  the  white  balls  may  be  drawn. 

5 
We  may  then  consider  —  as  the  likelihood  that,* if  a  ball  is 

drawn,  it  is  a  white  ball. 

The  drawing  of  a  white  hall  can  fail  to  occur  in  7  different 

ways ;  for  any  one  of  the  red  or  black  balls  may  be  drawn. 

7 
We  may  then  consider  —  as  the  likelihood  that,  if  a  ball  is 
^  12  ' 

drawn,  it  is  not  a  white  ball. 

634.  We  may  take  the  following  definition  for  the  term 
prohahility  : 

If  an  event  can  happen  in  a  different  ways,  and  fail  to  happen 
in  h  different  ways,  and  all  these  ways  are  equally  likely  to  occur j 

the  probability  of  the  happening  of  the  event  is r,  and  the 

h  CL  +  h 

probability  of  its  failing  is  . 

a-{-h 

"We  say  the  odds  are  a  to  6  in  favor  of  the  event,  if  a  is  greater  than  6, 
and  a  to  &  against  the  event,  if  a  is  less  than  h. 

It  follows  that  if  the  probability  of  the  happening  of  an 
event  is  p,  the  probability  of  its  failing  is  1—p. 

635.  Examples. 

1.    A  bag  contains  5  white,  4  red,  and  3  black  balls, 
(a)   If  3  balls  are  drawn,  what  is  the  probability  that  they 
are  all  white  ? 


<^^ 


PROBABILITY  437 

The  number  of  combinations  of  the  5  white  balls,  taken  2^ 
5.4.3 
at  a  time,  is  - — - — -  (§  626),  or  10 ;  that  is,  the  drawing  of  3 

white  balls  can  hagp^ninlO  different  ways. 

The  number  of  combinations  of  the  12  balls,  taken  3  at  a 

time,  is  ,  or  220;  that  is,  the  drawing  of  3  balls  can 

JL  •  Z  •  o 

occur  iiL^^  different  ways. 

Then,  the  probability  of  drawing  3  white  balls  is  -— ,  or  — . 

ZZO         ZIi 

(b)  If  6  balls  are  drawn,  what  is  the  probability  that  2  are 

white,  3  red,  and  |L  black  ? 

The  nuniber  of  combinations  of  the  5  white  balls,  taken  2  at 

5*4- 
a  time,  is ,  or  10 ;  the  number  of  combinations  of  the  4 

^•^  4.3.2 

red  balls,  taken  3  at  a  time,  is ,  or  4. 

'  '1.2.3' 

We  may  associate  together  any  one  of  the  10  combinations 
of  white  balls,  any  one  of  the  4  combinations  of  red  balls, 
and  any  one  of  the  3  black  balls ;  hence,  there  are  in  all 
10  X  4  X  3,  or  120,  different  combinations,  each  consisting  of 
2  white  balls,  3  red  balls,  and  1  black  ball. 

Also,  the  number  of  combinations  of  the  12  balls,  taken  6  at 
.    12  .  11  .  10  .  9  .  8  .  7        ciOA 
"  '^"''  ''     1.2.3.4.5.6    '  ''  ''^- 

Hence,  the  required  probability  is— —  ,  or  — . 

2.   A  bag  contains  30  tickets  numbered  1,  2,  3,  •••,  30. 

(a)   If  four  tickets  are  drawn,  what  is  the  chance  that  both 

1  and  2  will  be  among  them  ? 

The  number  of  combinations  of  the  28  tickets  numbered 

28  .  27 
3,  4,  ...,  30,  taken  2  at  a  time,  is       '      :   that  is,  there  are 

28  .  27 

'       different  ways  of  drawing  four  tickets,  two  of  which 

Jl  '  Z 

are  numbered  1  and  2. 


438  ADVANCED   COURSE   IN   ALGEBRA 

The  number  of  combinations  of  the  30  tickets,  taken  4  at  a 
, .        .    30  .  29  .  28  .  27 
^""^-^^      1.2.3.4     ' 

Hence,  the  probability  that,  if  four  tickets  are  drawn,  two 
of  them  will  be  1  and  2,  is 

28  .  27  .  30  •  29  .  28  .  27  ^   3.4   ^   2 
1.2     *       1.2.3.4         30 .  29     145* 

(6)  If  four  tickets  are  drawn,  what  is  the  chance  that  either 
1  or  2  will  be  among  them  ? 

Either  1  or  2  will  be  among  the  tickets  drawn,  unless  each 
ticket  drawn  bears  a  number  from  3  to  30  inclusive. 

The  number  of  combinations  of  the  28  tickets  numbered  3, 

4,  ...,  30,  taken  4  at  a  time,  is  ^^  '  ^'^  '  ^^  '  ^^. 
'      '      '  '  1.2.3.4 

The  number  of  combinations  of  the  30  tickets,  taken  4  at 

,.        .    30.29.28.27 

^^"^'-^^      1.2.3.4     • 

Hence,  the  probability  that  each  of  the  4  tickets  drawn  bears 

.       o        o  ^    OA  •     1     •       •    28  .  27  .  26  .  25        65 
a  number  from  3  to  30  inclusive,  is ,  or  — 

'      30  .  29  .  28  .  27        87 

Then,  the  probability  that  none  of  the  tickets  drawn  bears 

65  22 

a  number  from  3  to  30  inclusive,  is  1 (§  634),  or  — . 

87  87 

This  then  is  the  probability  that  either  1  or  2  will  be  among 

the  tickets  drawn. 

EXERCISE    105 

1.  A  bag  contains  6  white  balls,  5  red  balls,  and  4  black  balls ;  find 
the  probability  of  drawing  : 

(a)  One  black  ball.  (d)  Four  white  balls. 

(&)  Two  white  balls.  (e)  Two  balls  of  each  color, 

(c)  Three  red  balls.  (/)  Four  red  and  three  white  balls. 

{g)  Two  red,  five  white,  and  two  black  balls. 

2.  A  bag  contains  24  tickets  numbered  1,  2,  3,  ...,  24  ;  if  three  tickets 
are  drawn,  find  the  probability  : 

(a)  That  they  are  1,  2,  and  3. 

(6)  That  either  1,  2,  or  3  is  among  them. 


PROBABILITY  439 

3.  What  is  the  probability  of  throwing  not  more  than  5  in  a  single 
throw  with  two  dice  ? 

4.  What  is  the  probability  of  throwing  at  least  5  in  a  single  throw 
with  two  dice  ? 

5.  What  is  the  probability  of  throwing  10  in  a  single  throw  with  three 
dice? 

6.  If  six  persons  seat  themselves  at  random  at  a  round  table,  what  is 
the  probability  that  two  specified  persons  will  sit  together  ? 

7.  If  four  cards  are  drawn  from  a  pack,  what  is  the  probability  that 
they  are  of  the  same  suit  ? 

8.  There  are  8  books,  of  which  4  are  on  mathematics  and  3  on  science. 
If  the  books  are  placed  together  on  a  shelf,  what  is  the  probability  that 
the  mathematical  volumes,  and  also  the  scientific,  will  be  together  ? 

636.   Mutually  Exclusive  Events. 

If  an  event  can  happen  in  more  than  one  way,  and  if  it  happens  in  any 
of  the  ways,  cannot  at  the  same  time  happen  in  any  of  the  other  ways, 
these  various  ways  are  said  to  be  mutually  exclusive. 

If  an  event  can  happen  in  more  than  one  way,  and  these  ways 
are  mutually  exclusive,  the  probability  of  the  happening  of  the 
event  equals  the  sum  of  the  probabilities  of  its  happening  in  the 
separate  ways. 

Suppose  that  an  event  can  happen  in  a  certain  way  a  times 
out  of  b,  and  in  another  way  a'  times  out  of  b ;  all  these  ways 
being  equally  likely  to  occur. 

Also,  suppose  that  the  two  ways  in  which  th§  event  can 
happen  are  mutually  exclusive. 

Since  the  event  happens  a  +  a'  times  out  of  b,  the  probability 

of  its  happening  is  ^^^^  (§  634),  or  -  +  -• 
b  b      b 

But  -  is  the  probability  that  the  event  happens  in  the  first 

a' 
way,  and  —  the  probability  that  it  happens  in  the  second  way. 

Hence,  the  probability  that  it  happens  equals  the  sum  of 
the  probabilities  of  its  happening  in  the  separate  ways. 

In  like  manner,  the  theorem  may  be  proved  when  there  are 
more  than  two  ways  in  which  the  event  can  happen. 


440  ADVANCED   COURSE  IN   ALGEBRA 

637.  Examples. 

1.  Find  the  probability  of  throwing  4  in  a  single  throw  with 
two  dice. 

The  event  can  happen  in  two  ways ;  either  by  throwing  3 
and  1,  or  by  throwing  double-twos ;  and  these  ways  are  mutu- 
ally exclusive. 

Each  die  can  come  up  in  6  ways ;  and  hence  the  pair  can 
be  thrown  in  6  x  6,  or  36  ways. 

Of  these  different  throws,  two  will  be  3  and  1 ;  hence,  the 

2 
probability  of  throwing' 3  and  1  is  — 

36 

Again,  double-twos  can  be  thrown  in  only  one  way ;  hence,  the 
probability  of  throwing  double-twos  is  —  • 

2        11 

Therefore,  the  probability  of  throwing  4  is  — -  -f-  — -,  or  -—  • 

36     36        1^ 

This  example  can  be  solved  more  easily  by  the  method  of  §  635 ;  the 
above  method  is  given  simply  as  an  illustration  of  §  636. 

2.  A  bag  contains  four  $10  gold  pieces  and  six  silver  dol- 
lars. If  a  person  is  entitled  to  draw  two  coins  at  random, 
what  is  the  value  of  his  expectation  ? 

If  a  person  has  a  chance  of  winning  a  certain  sum  of  money,  the 
product  of  the  sum  by  the  probability  of  his  winning  it  is  called  his 
expectation. 

The  number  of  combinations  of  the  four  gold  pieces,  taken 

2  at  a  time,  is  — ^,  and  the  number  of  combinations  of  the  ten 

10  '9 
coins,  taken  2  at  a  time,  is  -i-^',  hence  the   probability  of 

4'3         2 
drawing  two  gold  coins  is  z-;—r,  or  --. 

10 '9        15 

Then  the  value  of  the  expectation,  so  far  as  it  depends  on 

2  -8 

the  drawing  of  two  gold  coins,  is  —  X  20,  or  -  dollars. 

6 '5        1 
The  probability  of  drawing  two  silver  coins  is  — -— ,  or  -; 

the  value  of  the  corresponding  expectation  is  -  dollars. 


PROBABILITY  441 

Again,  the  probability  of  drawing  a  gold  coin  and  a  silver 

coin  is  (6  ■  4)  -^      '      or  -— ;  the  value  of  the  corresponding  ex- 
1  '2         15 

8  88 

pectation  is  -—  x  11,  or  -—  dollars. 
15  15 

/8      2      88\ 
Hence,  the  value  of  the  expectation  is  f  -  +  -  +  —  J  dollars, 

or  $9.20.  ^^     ^     ^^^ 

EXERCISE   106 

1.  A  bag  contains  20  tickets  numbered  1,  2,  3,  •••,  20 ;  if  a  ticket  be 
drawn,  what  is  the  probability  that  its  number  is  a  multiple  of  3  or  7  ? 

2.  Find  the  probability  of  throwing  at  least  9  in  a  single  throw  with 
two  dice. 

3.  A  bag  contains  4  half-dollars  and  6  quarter-dollars.  If  a  person  is 
entitled  to  draw  a  single  coin,  find  the  value  of  his  expectation, 

4.  Find  the  probability  of  throwing  13  in  a  single  throw  with  three 
dice. 

5.  A  bag  contains  3  dimes,  4  five-cent  pieces,  and  2  twenty-cent 
pieces.  If  a  person  is  entitled  to  draw  two  coins,  what  is  the  value  of  his 
expectation  ? 

6.  Find  the  probahility  of  throwing  7  in  a  single  throw  with  four  dice. 

7.  A  bag  contains  7  gold  dollars,  and  5  five-dollar  gold  pieces.  If  a 
person  is  entitled  to  draw  four  coins,  what  is  the  value  of  his  expectation  ? 

8.  A  bag  contains  6  fifty-cent  pieces,  and  four  other  coins  which  have 
all  the  same  value.  If  a  person's  expectation  on  drawing  three  coins  is 
120|  cents,  find  the  value  of  each  of  the  unknown  coins. 


COMPOUND  EVENTS 

638.   Independent  Events. 

If  there  are  two  independent  events  whose  respective  probabilities 
are  known,  the  probability  that  both  will  happen  is  the  product  of 
their  separate  probabilities. 

Two  events  are  said  to  be  independent  when  the  occurrence  of  one  is 
not  affected  by  the  occurrence  of  the  other. 

Let  a  be  the  number  of  ways  in  which  the  first  event  can 
happen,  and  b  the  number  of  ways  in  which  it  can  fail;  all 
these  ways  being  equally  likely  to  occur. 


442       ADVANCED  COURSE  IN  ALGEBRA 

Also,  let  a'  be  the  number  of  ways  in  which  the  second  event 
can  happen,  and  b'  the  number  of  ways  in  which  it  can  fail ;  all 
these  ways  being  equally  likely  to  occur. 

We  may  associate  together  any  one  of  the  a-\-b  cases  in 
which  the  first  event  happens  or  fails,  and  any  one  of  the 
a'  -\-b'  cases  in  which  the  second  happens  or  fails ;  hence  there 
are  (a-f6)  (a'-\-b')  cases,  equally  likely  to  occur. 

In  aa'  of  these  cases  both  events  happen. 

Therefore,  the  probability  that  both  events  happen  is 

aa' 

(a  +  6)(a'  +  6')* 

But  — - —  is  the  probability  that  the  first  event  happens, 
a  +  6 

a' 

and  the  probability  that  the  second  happens. 

a  +6' 

Hence,  the  jDrobability  that  both  events  happen  is  the  product 
of  their  separate  probabilities. 

And  in  general,  if  pi,  P2,P3,  •••,  are  the  respective  probabilities 
of  any  number  of  independent  events,  the  probability  that  all 
the  events  happen  is  piP2P3*"« 

639.   Examples. 

1.  Find  the  probability  of  throwing  an  ace  in  the  first  only 
of  two  successive  throws  with  a  single  die. 

The  probability  of  throwing  an  ace  at  the  first  trial  is  ^* 

5 

The  probability  of  not  throwing  one  at  the  second  is  -• 

Hence,  the  probability  of  throwing  an  ace  in  the  first  only  of 

15        5 

two  successive  throws  is  -  X  -,  or  -— • 

6      6       oD 

2.  Find  the  probability  of  throwing  an  ace  at  least  once  in 
three  throws  with  a  single  die. 

There  will  be  an  ace  unless  there  are  three  failures. 

5 
The  probability  of  failing  at  the  first  trial  is  -;  and  this  is 

also  the  probability  of  failing  at  each  of  the  other  trials. 


PKOBABILITY  443 

Hence,  the  probability  that  there  will  be  three  failures  is 
5^55        125 

Then  the  probability  that  there  will  not  be  three  failures  is 

3.  A  bag  contains  5  red  balls,  4  white  balls,  and  3  black 
balls.  Three  balls  are  drawn  in  succession,  each  being  replaced 
before  the  next  is  drawn.  What  is  the  probability  that  the 
balls  drawn  are  one  of  each  color  ? 

The  probability  that  the  first  ball  is  red  is  — ;  the  proba- 

A  i 

bility  that  the  second  is  white  is  — ,  or  - ;  and  the  probability 

3         1 

that  the  third  is  black  is  — ,  or  -. 
'  12'       4 

Hence,  the  probability  of  drawing  a  red  ball,  a  white  ball, 

5      11          5 
and  a  black  ball,  in  this  assigned  order,  is  —  x  -  X  -,  or  . 

But  a  red  ball,  a  white  ball,  and  a  black  ball  may  be  drawn 
in  1^,  or  6  different  orders  (§  625)  ;  and  in  each  case  the  proba- 

bility  is  A. 

Then  by  §  636,  the  probability  of  drawing  a  red  ball,  a  white 
ball,  and  a  black  ball,  without  regard  to  the  order  in  which 

5  5 

they  are  drawn,  is  - —  x  6,  or  — . 

640.   Dependent  Events. 

TJie  probability  of  the  concurrent  happening  of  two  dependent 
events  is  the  probability  of  the  first,  multiplied  by  the  probability 
that  when  the  first  has  hapjjened  the  second  will  follow. 

Let  a  and  b  have  the  same  meanings  as  in  §  638. 

Also,  suppose  that,  after  the  first  event  has  happened,  a' 
represents  the  number  of  ways  in  which  the  second  will  follow, 
and  b'  the  number  of  ways  in  which  it  will  not  follow;  all 
these  ways  being  equally  likely  to  occur. 


444  ADVANCED  COURSE   IN   ALGEBRA 

Then  there  are  in  all  (a  +  b)  (a'  +  b')  cases,  equally  likely  to 
occur,  and  in  aa'  of  these  both  events  happen. 

Therefore,  the  probability  that  both  events  happen  is 


(a +  6)  (a' +  6') 

Hence,  the  probability  that  both  events  happen  is  the  proba- 
bility of  the  first,  multiplied  by  the  probability  that  when  it 
has  happened  the  second  will  follow. 

And  in  general,  if  there  are  any  number  of  dependent  events 
such  that  pi  is  the  probability  of  the  first,  p2  the  probability 
that  when  the  first  has  happened  the  second  will  follow,  pg  the 
probability  that  when  the  first  and  second  have,  happened 
the  third  will  follow,  and  so  on,  then  the  probability  that  all 
the  events  happen  is  P1P2P3  ••*• 

641.    Examples. 

1.  Solve  Ex.  3,  §  639,  if  the  balls  are  not  replaced  after  being 
drawn. 

The  probability  that  the  first  ball  is  red  is  —  ;  the  probabil- 
ity that  the  second  is  white  is  — ;  and  the  probability  that  the 

3 

third  is  black  is  — . 
10 

Hence,  the  probability  of  drawing  a  red  ball,  a  white  ball, 

5       4       3 
and  a  black  ball,  in  this  assigned  order,  is  —  x  —  X  —  • 

x^      11      10 

But  the  balls  may  be  drawn  in  [3,  or  6  different  orders. 

Therefore,  the  probability  of  drawing  a  red  ball,  a  white 

ball,  and  a  black  ball,  without  regard  to  the  order  in  which 

they  are  drawn,  is  —-  x  -—  x  — :  X  6,  or  —  • 
^         pjj        '      12     11      10       '11 

2.  An  urn  contains  5  white  balls  and  3  black  balls ;  another 
contains  4  white  balls  and  7  black  balls.  What  is  the  proba- 
bility of  obtaining  a  white  ball  by  a  single  drawing  from  one 
of  the  urns  taken  at  random  ? 


PROBABILITY  445 

Since  the  urns  are  equally  likely  to  be  taken,  the  proba- 
bility of  taking  the  first  urn  is  -;  and  the  probability  of  then 

Li 

5 

drawing  a  white  ball  from  it  is  -• 

8 

Hence,  the  probability  of  obtaining  a  white  ball  from  the 

15        5 

first  urn  is  -  X  -,  or  — • 

In  like  manner,  the  probability  of  obtaining  a  white  ball 

from  the  second  urn  is  -  x  — ,  or  — 
Z      11        11 

5       2         87 

Hence,  the  required  probability  is 1 ,  or  - — • 

'  ^  ^  -^      16     11       176 

642.  Given  the  probability  of  the  happening  of  an  event  in 
one  trial,  to  find  the  probability  of  its  happening  exactly  r  times 
in  n  trials. 

Let  p  be  the  probability  of  the  happening  of  the  event  in 
one  trial. 

Then  1  —  p  is  the  probability  of  its  failing  (§  634). 

JThe  probability  that  the  event  will  happen  in  each  of  the 
first  r  trials,  and  fail  in  each  of  the  remaining  n  —  r  trials,  is 

But  the  number  of  ways  in  which  the  event  may  happen 
exactly  r  times  in  n  trials  is  equal  to  the  number  of  combina- 
tions of  n  things  taken  r  at  a  time,  or 

n(n-l).-(n-r  +  l)  ^^  ^26). 

\L 

Hentie,  the  probability  that  the  event  will  happen  exactly 
r  times  in  n  trials  is 

\l 

Ex.  What  is  the  probability  of  throwing  exactly  three  aces 
in  five  throws  with  a  single  die  ? 


446       ADVANCED  COURSE  IN  ALGEBRA 


Here,  ^~'a*  r  =  3,  n  =  5,  and  n  —  r  -h  1  =  3. 

Substituting  in  (1),  the  required  probability  is 

'•^•'xr^Yxr^Yor^^^ 


1-2.3     \(oj      \6J'      3888 

643.  It  follows  from  §  642  that,  if  the  probability  of  the 
happening  of  the  event  in  one  trial  is  p,  the  probability  of  its 
failing  exactly  r  times  in  oi  trials  is 

\l 

644.  Given  the  probability  of  the  happening  of  an  event  in  07ie 
trial,  to  find  the  probability  of  its  happening  at  least  r  times  in  n 
trials. 

The  event  happens  at  least  r  times  if  it  happens  exactly  n 
times,  or  fails  exactly  once^,  twice,  •••,  n  —  r  times. 

Then  the  probability  ^at  it  happens  at  least  r  times  equals 
the  sum  of  the  probabilities  of  its  happening  exactly  n  times, 
or  failing  exactly  once,  twice,  •••,  n  —  r  times. 

By  §§  642,  643,  the  required  probability  is 

\n~r 

Ex.  What  is  the  probability  of  throwing  at  least  three  aces 
in  five  throws  with  a  single  die  ? 

Here,  P  =  -^,  r  =  3,  n  =  5',  then  the  required  probability  is 


67  \6J       \6       1-2      \6         \6r       648 


EXERCISE    107 

1.  Find  the  probability  of  throwing  exactly  four  sixes  in  six  throws 
with  a  single  die. 

2.  Find  the  probability  of  throwing  at  least  four  doublets  in  six  throws 
with  a  pair  of  dice. 


PROBABILITY  447 

3.  A  purse  contains  5  dollars  and  7  five-cent  pieces,  and  another 
3  dollars  and  12  five-cent  pieces.  Find  the  probability  of  obtaining  a 
dollar  by  drawing  a  single  coin  from  one  of  the  purses  taken  at  random. 

4.  If  a  coin  is  tossed  eight  times,  what  is  the  probability  that  the  head 
will  turn  up  at  least  live  times  ? 

5.  A  bag  contains  5  white  and  3  black  balls.  If  4  balls  are  drawn  and 
not  replaced,  what  is  the  probability  that  the  balls  drawn  are  alternately 
of  different  colors  ? 

6.  What  is  the  probability  of  throwing  10  with  a  pair  of  dice  exactly 
three  times  in  four  trials  ? 

2 

7.  The  probability  of  a  certain  event  is  - ,  and  of  another  independent 

a  ' 

of  the  first  — .    Find  the  probability  that  one  at  least  of  the  events  will 
happen. 

8.  If  two  coins  are  tossed  up  five  times,  find  the  probability  that  there 
will  be  five  heads  and  five  tails. 

9.  Each  of  four  persons  draws  a  card  from  a  pack.  Find  the  proba- 
bility that  there  will  be  one  of  each  suit. 

10.  A,  B,  C.  and  D  throw  a  die  in  succession,  in  the  order  named, 
until  one  throws  an  ace.  Find  their  respective  chances  of  throwing  an 
ace  at  the  first  trial.  » 

11.  A  bag  contains  three  white  and  six  black  balls.  A  person  draws 
three  balls,  the  balls  when  drawn  not  being  replaced.  What  is  the 
probability  of  drawing  a  white  ball  ? 

12.  A  person  has  four  tickets  in  a  lottery  in  which  there  are  three  prizes 
and  seven  blanks.     Find  his  chance  of  drawing  a  prize. 

13.  A  box  contains  ten  counters  numbered  1,  2,  3,  ...,  10.  After  one 
is  drawn,  it  is  put  back,  and  the  process  is  repeated  indefinitely.  Find- 
the  probability  that  No.  1  will  be  drawn  in  four  trials. 

14.  A  bag  contains  six  balls.  A  person  takes  one  out,  and  replaces  it. 
After  he  has  done  this  six  times,  find  the  probability  that  he  has  had  in 
his  hand  every  ball  in  the  bag. 

15.  In  a  series  of  games,  the  probability  that  the  winner  of  any  game 

2 
wi^  the  next  game  is  -.     Find  the  probability  that  the  winner  of  the 

o 
first  game  wins  three  or  more  of  the  next  four. 

16.  A  bag  contains  three  tickets  numbered  1,  2,  3.  A  ticket  is  drawn, 
and  replaced.  After  this  has  been  done  four  times,  what  is  the  probability 
that  the  sum  of  the  numbers  drawn  is  even  ? 


448  ADVANCED  COURSE  IN  ALGEBRA 

17.  A  purse  contains  a  silver  dollar  and  four  dimes  ;  another  contains 
five  dimes.  Four  coins  are  taken  from  the  former  and  put  in  the  latter  ; 
and  then  four  coins  are  taken  from  the  latter  and  put  in  the  former.  Find 
the  probability  that  the  dollar  is  still  in  the  first  purse. 

18.  A  and  B,  with  six  others,  draw  lots  for  partners,  and  play  four  two- 
handed  games,  all  the  players  being  of  equal  skill.  The  four  winners  draw 
lots  for  partners,  and  play  two  games,  and  the  winners  in  these  games  play 
a  final  game.    Find  the  probability  that  A  and  B  have  played  together. 

19.  If  four  whole  numbers  taken  at  random  be  multiplied  together,  find 
the  chance  that  the  last  digit  in  the  product  is  1,  3,  7,  or  9. 

20.  An  urn  contains  3  white  and  3  black  balls,  and  another  4  white  and 
4  black  balls.  A  ball  is  taken  from  one  and  put  in  the  other.  If  a  ball 
be  drawn  from  one  of  the  urns  chosen  at  a  random,  what  is  the  probability 
that  it  is  white  ? 


CONTINUED   FRACTIONS  449 


XXXIII.    CONTINUED  FRACTIONS 

645.   A  Continued  Fraction  is  an  expression  of  the  form 

h  h        d 

a-\ 5 — ;    or,  a-\- 


c-{-  e-\ 

as  it  is  usually  written. 

We  shall  consider  in  the  present  work  only  continued  fractions  of  the 
form  1         1 


&  +  c  +  ...  ' 

where  each  numerator  is  unity,  a  a  positive  integer  or  0,  and  each  of  the 
numbers  &,  c,  •..,  a  positive  Integer. 

646.  A  terminating  continued  fraction  is  one  in  which  the 
number  of  denominators  is  finite ;  as, 

,111 

h-\-  c-\-  d 

An  iyijinite  continued  fraction  is  one  in  which  the  number  of 
denominators  is  indefinitely  great. 

647.  In  the  continued  fraction 

,11  1 

Ct2  +    «3  +    <^4  +  •  •  • 

«!  is  called  the^?'s^  convergent; 
tti  H —  the  second  convergent; 

ai  -\ the  third  convergent;  and  so  on. 

If  ai  =  0,  as  in  the  continued  fraction 
11  1 


«2  +  «3  +  a4  +  ••• 
then  0  is  considered  the  first  convergent. 

648.  Any  ordinary  fraction  in  its  loivest  terms  may  he  con- 
verted into  a  terminating  continued  fraction. 


450       ADVANCED  COURSE  IN  ALGEBKA 


Let  the  given  fraction  be  -,  where  a  and  h  are  prime  to  each 
other. 

Divide  a  by  h,  and  let  aj  denote  the  quotient  and  h^  the  re- 
mainder; then, 

a  ,  hi  ,1 

h  h  h_ 

Divide  b  by  bi,  and  let  a2  denote  the  quotient  and  &2  the  re- 
mainder; then, 

a  ,       1  ,1 


«2  +  7^  ^2+- 

&2 

Again,  divide  6i  by  &2?  ^-nd  let  ag  denote  the  quotient  and  63 
the  remainder ;  then, 

a  ,  1  ,1 

-  =  a,  +  -—^  =  a,  +  -—^-. 

a2-\ ajH T- 

The  process  is  the  same  as  that  of  finding  the  H.  C.  F.  of  a 
and  b  (§  188)  ;  and  since  a  and  b  are  prime  to  each  other,  we 
must  eventually  obtain  a  remainder  unity,  at  which  point  the 
operation  terminates. 

Hence,  any  ordinary  fraction  in  its  lowest  terms  can  be  con- 
verted into  a  terminating  continued  fraction. 

Ex.     Convert  —  into  a  continued  fraction. 
23 

23)62(2  =  ai 
46 

16)23(1  =  a^ 
16 
7)16(2  =  a3 
14 
2)7(3  =  a4 
6 
1 


CONTINUED   FRACTIONS  451 

Therefore,  ^  =  2  +  —  -r^-i-^- 

'23  1+2+34-2 

649.  A  quadratic  surd  (§  368)  may  be  converted  into  an  infinite 
continued  fraction. 

Ex.     Convert  V6  into  a  continued  fraction. 
The  greatest  integer  in  V6  is  2 ;  we  then  write 
V6  =  2  +  (V6-2). 

Keducing  V6  —  2  to  an  equivalent  fraction  with  a  rational 
numerator  (§  387),  we  have 

V6+2  V6+2 

2 

The  greatest  integer  in  ^  is  2  ;  we  then  write 

Z 

V6  +  2^^,  I  V^-^^2  I  (V^-2)(V6  +  2)^^  J        1 
2  2  .  .  2(V6  +  2)  V6  +  2* 

Substituting  in  (1),  V6  =  2  H ^ (2) 

2  +  -J— 
V6  +  2 

The  greatest  integer  in  V6  +  2  is  4 ;  we  then  write 

V6  +  2  =  4  +  (V6-2)  =  4  +  (^^^^:iMV6_±2} 

V6  +  2 
=  4+-^  =  4+       ^ 


Substituting  in  (2),  we  have\ 

V6  =  2  + Lj 

'  + 4- 

/  V6  +  2 


452  ADVANCED   COURSE   IN  ALGEBRA 


The  steps  now  recur,  and  we  have 
V6=V'     1111 


2  +  4^-2  +  4  +  ... 

An  infinite   continued  fraction  in  which  the  denominators  recur  is 
called  a  periodic  continued  fraction. 

650.   A  periodic  continued  fraction  may  always  he  expressed 
as  an  irrational  7iumber. 

Ill        1 

Ex.    Express  - —  - —  - — as  an  irrational  number. 

1  +  3  +  1  +  3+.-. 

Let  X  denote  the  value  of  the  fraction  ;  then, 

^^    1       1     ^     3  +  a;     ^S-\-x 
1  +  3  +  a;     3  +  a;+l      4  + a;' 

Clearing  of  fractions, 

4 ic  +  aj^  =  3  +  ic,  or  x^-\-Sx  =  S. 


Solving  this  equation,  x  =  -^+^^  +  12 ^  -3  +  V21. 
It  is  evident  that  the  positive  sign  must  be  taken  before  the  radical. 

PROPERTIES  OF  CONVERGENTS 

651.    In  §§  652  to  657,  inclusive,  we  shall  suppose  the  con- 
tinued fraction  to  be 

«.+  1111 


^2+  %+        a„+  a„  +  iH 

And  we  shall  let  j?^  denote  the  numerator,  and  q,.th.e  denomi- 
nator, of  the  rth  convergent  (§  647),  when  expressed  in  its 
simplest  form. 

652.  To  determine  the  laio  of  formation  of  the  successive 
convergents. 

The  first  convergent  is  %. 

The  second  is  aj  +  i  =  %^^2+J:. 


// 


CONTINUED   FRACTIONS  453 

Thethirdis     0,  +  ^-  l=a,+  -^i-  =  ^lMi±^L±±3. 

^2+   «3  a2«3  +  l  a2«3  +  l 

The  third  convergent  may  be  written  in  the  form 
(aia2  4-l)^.3  +  cti, 

a2«3  + 1        ' 
in  which  we  observe  that : 

I.  Tlie  numerator  equals  the  numerator  of  the  preceding  con- 
vergent, multiplied  by  the  last  denominator  taken,  plus  the  numera- 
tor of  the  convergent  next  hut  one  preceding. 

II.  The  denomiyiator  equals  the  denominator  of  the  preceding 
convergent,  multiplied  by  the  last  denominator  taken,  plus  the 
denominator  of  the  convergent  next  but  one  preceding. 

We  will  now  prove  by  Mathematical  Induction  that  the  above 
laws  hold  for  all  convergents  after  the  second,  when  expressed 
in  their  simplest  forms. 

Assume  that  the  laws  hold  for  all  convergents  as  far  as  the 
?ith  inclusive. 

The  nth  convergent  is^  =  aiH •••  — . 

qn  ^2+  a3+         «« 

Then,  since  the  last  denominator  is  a„,  we  have 

JPn  =  a«2)n-i+i^«-2,  andg„  =  a„g„_i  +  g„_2.    .  (1) 

Whence,  ^  =  %Pn-i+Pn-2^  (2) 

The  (n  +  l)th  convergent  is 

,11  11 

«2+    «3+  ««+   ««+l 

which  differs  from  the  nth  only  in  having  a„4- ,  or  ^»^"+i"^  , 

in  place  of  a^.  ^"^^  ^«+i 

Substituting  ^n^^+i-hl  ^^^  ^^  -^  ^2),  we  have 

^nCtn+l  +  l 

Pn-l  -1-Pn-2 

Pn+1  __  <*n+l ^ 

Qn+i      M«±L±io      -i-^ 

^n-1  +  Qn-2 


454  ADVANCED   COURSE   IN   ALGEBRA 


Then, 


9n+l         Cln+1  («n  Qn-l  +  Qn-z)  +  ?„-! 


««+l2^«+i?n- 


^  by  (1).  (3) 


It  is  evident  that  the  second  member  of  (3)  is  the  simplest 
form  of  the  (n  +  l)th  convergent,  and  therefore 

Pn+l  =  an+iPn+Pn-l,    ^nd    Qn+l  =  Ctn-^lQn  +  Qn-V 

These  results  are  in  accordance  with  laws  I  and  II. 

Hence,  if  the  laws  hold  for  all  convergents  as  far  as  the  nth 
inclusive,  they  also  hold  for  the  (n  -\-  l)th. 

But  we  know  that  they  hold  for  the  third  convergent,  and 
hence  they  hold  for  the  fourth ;  and  since  they  hold  as  far  as 
the  fourth,  they  also  hold  for  the  fifth ;  and  so  on. 

Hence,  the  laws  hold  for  all  convergents  after  the  second. 

Ex.   Find  the  first  five  convergents  of 

1  +  ^-1.-1-       1 


1+ 2+3+4+ ... 
The  first  convergent  is  1,  and  the  second  is  1  + 1,  or  2. 
Then,  by  aid  of  the  laws  just  proved, 

the  third  is 


the  fourth  is 
the  fifth  is 


2. 

2  +  1 

5 

1. 

^  +  1 

"3' 

5. 

3  +  2 

17 

3. 

3  +  1 

~10' 

17 

.4  +  5 

73 

10 

.4  +  3 

43 

653.    The  difference  between  two  consecutive  convergents   _ 
id  ^n±l  is  -1 

Qn+l  qnQn+l 

The  difference  between  the  first  and  second  convergents  is 

^     as)       ^  ~  0^2 
Thus  the  theoi-em  holds  for  the  first  and  second  convergents. 


continup:d  fractions  455 

Assume  that  it  holds  for  the  nth.  and  (71  +  l)th  convergents. 
That  is,    ^~-^'  =  — -— ,  ov  p^q^+^r^p,+^q^==l.  (1) 

Then, 7^-  =  -^ ."T-^^ — 4777     (§^^2) 

gn+l(a«+2gn+l  +  5«) 
gn+ign+2  ^  ^         gn+l9«+2 

Hence,  if  the  theorem  holds  for  any  pair  of  consecutive  con- 
vergents, it  also  holds  for  the  next  pair. 

But  we  know  that  it  holds  for  the  first  and  second  conver- 
gents, and  hence  it  also  holds  for  the  second  and  third ;  and 
since  it  holds  for  the  second  and  third,  it  also  holds  for  the 
third  and  fourth ;  and  so  on. 

Therefore,  the  theorem  holds  universally. 

As  an  example  of  the  theorem,  the  difference  Ipetween  the  fourth  and 
fifth  convergents,  in  the  example  in  §  652,  is 

17      73  ^  731  -  730  ^       1 
10     43       10x43       10x43* 

654.  It  follows  from  §  653  that  p^  and  q^  can  have  no  com- 
mon divisor  except  unity;  for  if  they  had,  it  would  be  a  divisor 
of  i>n^n+i~i5n+i^»j  0^  uuity,  wMch  is  impossible. 

Therefore,  all  convergents  formed  in  accordance  with  the 
laws  of  §  652  are  in  their  lowest  terms. 

655.  Tlie  even  convergents  are  greater,  and  the  odd  conver- 
gents less,  than  the  fraction  itself. 

I.  The  first  convergent,  ai,  is  less  than  the  fraction  itself, 

since  is  omitted. 

a2H 

II.  The  second,  a^  -\ — ,  is  greater,  because  its  denominator  Oj 

^  <^2 

is  less  than  ctg  -\ — — ,  the  denominator  of  the  fraction. 

c^s  ~r  *  *  * 


456  ADVANCED   COURSE  IN  ALGEBRA 

III.    The   third,   ai  -\ — ,   is   less,   because,   by   II,  the 

-1  ^a  ~f"  %  11 

denominator   00+—    is    greater    than    a^ -] ,   the 

as         ^  -      a3+a,  H 

denominator  of  the  fraction;   and  so  on. 

Hence,  the  first,  third,   •••,  convergents  are  less,  and  the 
second,  fourth,  •••,  convergents  greater  than  the  fraction  itself. 

656.   Any  convergent  is  nearer  than  the  preceding  convergent 
to  the  value  of  the  /inaction  itself. 

By   §652,  P^^<^n.^Pn..+Pn^ 

The  fraction  itself  is  obtained  from  its  (n  -f  2)th  convergent 

by  putting  a^+2  H j- —  in  place  of  a„+2- 

Hence,  denoting  the  value  of  the  fraction  by  x,  we  have 
1 


««4-2H-r 


«n+3+"- 

where  m  stands  for  a„+2  ' 


P'^^+P"  ™p„,,+p„ 


Now,     X  • 


^'m{Pn+iqn'^Pnqn+l} 

qn('^qn+i  +  qn) 

(§653).  •  (1) 


Also,  X 


qn('^qn+i  +  qn) 

Pn+l  ^  '^Pn+1  +Pn  ^Pn+1 

qn+1     mq^^^  +  qn^  qn+i 

Pnqn+y^Pn+iqn 


^n+l(^g«+l  +  qn)         qn+l(inqn+l  +  ^n) 


(2) 


Since  a^+2  is  a  positive  integer,  a„^2  H r —  is  >  1 ;  that 

is,mis>l.  «n+3  +  - 

And  since  g„+i  =  a^+,q^  +  $n-i  (§  652),  q^^^  is  >  g«. 


CONTINUED   FRACTIONS  457 

Therefore,  the  fraction  (2)  is  less  than  the  fraction  (1),  for 
it  has  a  smaller  numerator  and  a  greater  denominator. 

Hence,  the  (n  +  l)th  convergent  is  nearer  than  the  wth  to 
the  value  of  the  fraction  itself. 

^    657.    To  determine  limits  to  the  error  made  in  taking  the  nth 
convergent  for  the  fraction  itself 

With  the  notation  of  §  &^Q>,  the  difference  between  the 
fraction  itself  and  its  nth  convergent  is 

m.  1 


qni^Qn+l  +  qn) 


'f^^  +  S 


(1) 


Since  m  is   >1   (§  656),  the  denominator  9,/9„+i  +  — ) 
<?,(9„«  +  9-.)-  ^  '^^ 

The  denominator  is  also  >  qnQn+i- 

Hence,  the  fraction  (1)  is  >  — ■. — - ,  and  < 


That  is,  the  error  made  in  taking  the  ?ith  convergent  for  the 
fraction  itself  lies  between 

1  and      1 


qniQn+l  +  qn)  qnqn+l 

As  an  example  of  the  above  theorem,  the  error  made  in  taking  the 
fourth  convergent  for  the  fraction  itself,  in  the  example  in  §  652,  lies 
between                      -                         ill 
and ,  or  —  and 


10(43  +  10)  10x43'        530  4^0 


EXERCISE    108 

Convert  each  of  the  following  into  a  continued  fraction,  and  find  in 
each  case  the  first  five  convergents.  • 

J     118  3     145  6     118  ^      n^ 

■     91  '  *    612'  '    67l'  ■    1561' 

2.    2o3.  4.   5.83.  6.    ^.  8.    ^. 

179  611  5151 

Convert  each  of  the  following  into  a  continued  fraction,  find  in  each 
case  the  first  four  convergents,  and  determine  limits  to  the  error  made  in 
taking  the  third  convergent  for  the  fraction  itself, 


458  ADVANCED   COURSE  IN   ALGEBRA 


9. 

\/26. 

JO. 

V37. 

11. 

a/8. 

12. 

Vie. 

13. 

4- 

14. 

V27. 

15. 

v/63. 

16. 

Vl9. 

17 

1  +\/46 

18.  V23. 

19.  2\/7. 

20.  3\/l4. 


Express  each  of  the  following  in  the  form  of  a  surd 
1111 


21 


4+3+4+3+ 
1111 


2+5+2+5+ 

25.    a  + 


1 


23. 

4  + 

1         1 

2  +  2+  ... 

24 

1  + 

1       1       1 

1 

1+6+1+6 

+  ••• 

1 

1 

a  +  2a  +  a  +  2a  +  --. 

26.  The  sidereal  year  is  approximately  365.25636  days ;  express  the 
excess  above  365  days  as  a  continued  fraction,  and  find  its  first  four 
convergents. 

27.  A  kilometer  is  approximately  .62138  mile  ;  express  this  decimal 
as  a  continued  fraction,  find  its  fifth  convergent,  and  determine  limits  to 
the  error  made  in  taking  this  convergent  for  the  fraction  itself. 

28.  A  meter  is  approximately  1.09.863  yards  ;  express  this  decimal  as  ^ 
continued  fraction,  find  its  sixth  convergent,  and  determine  limits  to  the 
error  made  in  taking  this  convergent  for  the  fraction  itself. 

29.  Express  the  greatest  root  of  the  equation 

2  ic2  -  10  X  =  -  5 
as  a  continued  fraction,  and  find  the  first  five  convergents. 

Convert  each  of  the  following  into  a  continued  fraction,  and  find  in 
each  case  the  first  four  convergents : 

30.  V74.  31.   -^-  32.    \/a2  +  a.  33.    V57. 

V55 


SUMMATION  OF   SERIES  459 


XXXIV.    SUMMATION  OP  SERIES 

658.  The  Summation  of  an  infinite  literal  series  is  the  pro- 
cess of  finding  an  expression  frotn  which  the  series  may  be 
developed. 

In  §  630,  we  gave  a  method  for  finding  the  sum  of  an  infinite  geometric 

series. 

RECURRING  SfRIES 

659.  Consider  the  infinite  series 

1  +  2  a?  +  3  x'^  +  4  a;3  +  5  a;*  + . . . . 

Here,  (3  a^  -  2  x(2  x)  +  x\l)  =  0, 

(4  a^)  -  2  x(3  x')  -\-  x\2  x)  =  0,  etc. 

That  is,  any  three  consecutive  terms,  as  for  example  2x,  Sx^, 
and  4  a?^,  are  so  related  that  the  third,  minus  2  x  times  the 
second,  plus  a^  times  the  first,  equals  0. 

660.  A  Recurring  Series  is  an  infinite  series  of  the  form 

tto  +  OriOS  +  a^x^  -\ , 

where  any  r  + 1  consecutive  terms,  as  for  example 

are  so  related  that 

a„a;«+M«n-ia?"-')  +  gaj2(a„_2a:«-2)+  ...  -f  sa;'-(a„_,af»-'-)  =  0  ; 
p,  q,  '",  s  being  constants. 

The  above  recurring  series  is  said  to  be  of  the  rth  order,  and 
the  expression  i +^^  +  3^.  + ... +  ,^. 

is  called  its  scale  of  relation. 

The  recurring  series  of  §  G59  is  of  the  seconji  order,  and  its  scale  of 
relation  is  1  —  2  x  +  x^.      

An  infinite  geometric  series  is  a  recurring  series  of    the  first  order. 


460  ADVANCED   COURSE  IN   ALGEBRA 

Thus,  in  the  infinite  geometric  series 

1  +a;  +  cc2  +  x3+  •••, 
any  two  consecutive  terms,  as  for  example  x^   and   x^,  are  so  related 
that  (x^)  —  x(x^)  =  0  ;  and  the  scale  of  relation  is  1  —  ic. 

661.    To  find  the  scale  of  relation  of  a  recurring  series. 

If  the  series  is  of  the  first  order,  the  scale  of  relation  may  be 
found  by  dividing  any  term  by  the  preceding  term,  and  sub- 
tracting the  result  from  1. 

If  the  series  is  of  the  second  order,  ao,  a^,  a^,  a^,  •••,  its  con- 
secutive coefficients,  and  l-\-px-\-qx^  its  scale  of  relation,  we 
shall  have  (a,+pa,  +  qao  =  0, 

ttg  +pa2  +  qai  =  0 ;  ^  ^ 

from  which  p  and  q  may  be  determined. 

If  the  series  is  of  the  third  order,  aQ,  a^,  a^,  a^,  a^  a^,  •••,  its 
consecutive  coefficients,  and  1 -{- px -{- qx^ -{-  ra^  its  scale  of  rela- 
tion, we  shall  have 

'  03  -f-  pa2  +  qai  -{-  rao  =  0, 
•  a^-{-pas  +  qa2  +  rai  =  0, 
.  as +pa4  +  (/a3  H- m2  =  0 ; 

from  which  p,  q,  and  r  may  be  determined. 

It  is  evident  from  the  above*  that  the  scale  of  relation  of  a 
recurring  series  of  the  rth  order  may  be  determined  when  any 
2r  consecutive  terms  are  given. 

To  ascertain  the  order  of  a  series,  we  may  first  make  trial  of 
a  scale  of  relation  of  three  terms ;  if  the  result  does  not  agree 
with  the  series,  try  a  scale  of  four  terms,  five  terms,  and  so  on 
until  the  correct  scale  of  relation  is  found. 

If  the  series  is  assumed  to  be  of  too  high  an  order,  the  equa- 
tions corresponding  to  the  assumed  scale  will  not  be  indepen- 
dent.    (Compare  §  269.) 

662=  To  find  the  sum  of  a  recurring  series  when  its  scale  of 
relation  is  known. 

Let  1  -\-px-{-  qxP  be  the  scale  of  relation  of  the  series 


SUMMATION   OF  SERIES  461 

Denoting  the  sum  of  the  first  n  terms  by  S,^,  we  have 

Then,       pxS^  =  pa^x  -i-pa^of  -\ hpo^„-2^""^  +  i>«„_i«", 

and  qx^S„  =  qa(fic^  H h  qa^^^sif-^  +  qa„_2^  +  go^«-li»"+^ 

Adding  these  equations,  and  remembering  that,  by  virtue  of 
the  scale  of  relation, 

the  coefficients  of  x^,  oif,  •••,  aj"~^  become  0,  and  we  have 

Sn(l+px-{-qx^ 

=  Oo  +  («i  +i>«o)  ^  +  (i>««-i  +  gotn-2)  a?"  +  ga„_ia;"+^ 
Whence, 

o  _ gp  +  («i  +pao) a;  +  (pan-i  +  g<^n-2) a^"  +  qan-i^'"'^'^ .      n\ 
l+px  +  qaf  '       ^^^ 

which  is  a  formula  for  the  sum  of  the  first  n  terms  of  a  recurring 
series  of  the  second  order. 

If  X  is  so  taken  that  the  given  series  is  convergent,  the 
expression  ^^^^^  j^  ^^^_^)  ^n  ^  ga„_ia;"+i 

approaches  the  limit  0  when  n  is  indefinitely  increased  (§  542), 
and  (1)  becomes         ^  ^  a, -^  (a, -^  pa,)  x ,  .^. 

l-{-px  +  qa^    '  ^  ^ 

which  is  a  formula  for  the  value  (§  540)  of  a  recurring  series 
of  the  second  order. 

If  q  =  Oy   the   series   is   of  the   first  order,  and   therefore 
Oh  +  mo  =  0 ;  whence,  ^ 

which  is  a  formula  for  the  value  of  a  recurring  series  of  the 
first  order.     (Compare  §  530.) 

In  like  manner,  we  shall  find  the  formula 

^  _  ftp  +  («!  -j-paj))  X  +  («o  +  pai  4-  qoo)  x^  .^. 

l-\-px-{-  qx^  H-  rx^ 

for  the  value  of  a  recurring  series  of  the  third  order. 


462  ADVANCED   COURSE   IN   ALGEBRA 

It  will  be  observed,  in  formulae  (1),  (3),  and  (4),  that  the  denominator 
is  the  scale  of  relation.  

A  recurring  series  is  formed  by  the  expansion,  in  an  infinite  series,  of  a 
fraction,  called  the  generating  fraction.  The  operation  of  summation 
reproduces  the  fraction ;  the  process  being  just  the  reverse  of  that  of  §  564. 

Ex.     Find  the  sum  of  the  series 

2  +  a;  +  5  0^2  +  7  ic^  + 1 7  a;4  + . . . . 

To  determine  the  scale  of  relation,  we  first  assume  the  series 
to  be  of  the  second  order  (§  661). 

Substituting  aQ  =  2,  ai  =  1,  ag  =  5,  Wg  =  1,  in  (1),  §  661, 
|5+    i>  +  2g  =  0, 
l7  +  5i9+    g  =  o. 

Solving  these  equations,  p  =  —  1,  g  =  —  2. 

To  ascertain  if  1  —  a;  —  2  a;^  is  the  correct  scale  of  relation, 
consider  the  fifth  term. 

Since  17  a;^  +  (-  x)  (7  aj^)  +  (-  2 x")  (5  x^)  is  0,  it  follows  that 
1  —  a;  —  2  a?^  is  the  correct  scale. 

Substituting  the  values  of  do,  ai,  p,  and  q  in  (2), 

o^2  +  (l-2)a;^       2  -  a; 
l-x-2x'       l_a;-2a;2' 

The  result  may  be  verified  by  expansion. 


The  series  expresses  the  value  of  the  fraction  only  for  such  values  of  x 
as  make  the  series  convergent. 

To  find  for  what  values  of  x  the  given  series  is  convergent,  we  proceed 
as  in  §  572  ;  we  find  by  the  method  of  §  567, 

2-x        _       1        ,1 


l-x-2a;2     \-2x     \  +  x 

=  (1  +  2  cc  +  2%2 +...)  +  (!  _a;  +  cc2 ). 

The  nth  term  of  the  given  series  is  [2'*-i  +  (—  1)*'~^]  x"-i. 
The  ratio  of  the  (n  +  l)th  term  to  the  nth  term  is 


or  ^-— 


[2»+(-  l)^]a;^ 


SUMMATION   OF   SERIES  463 

This  approaches  the  limit  2  x,  when  n  is  indefinitely  increased. 
Then,  the  series  is  convergent  if  x  is  numerically  <-  (§  555,  I.). 

EXERCISE  109 

In  each  of  the  following  find  the  generating  fraction,  and  the  expres- 
sion for  the  nth  term,  and  determine  for  what  values  of  x  the  series  is 
convergent :     - 

1.  4  -  X  +  7  ic2  -  5  a;3  +  19  ic*  +  —. 

2.  1  -  13  x  -  23  x2  -  85  5c3  -  239  x*  +  .... 

3.  1  +  5  X  +  21  x2  +  85  a:3  +  341  x*  +  .... 

4.  5  ~  13  X  +  35  x2  -  97  x3  +  275  x*  +  .... 
6.   3  +  10  X  +  36  x2  +  136  x^  +  528  x*  +  .... 

6.  1  -  2  X  +  x2  +  22  x3  -  191  x*  +  .-. 

7.  3  +  X  +  33  x2  +  109  x3  +  657  x*  +  •••. 

8.  1  +  31  X  -  19  x2  +  391  x3  -  619  x*  +  .... 

In  each  of  the  following  find  the  generating  fraction,  and  continue  the 
series  to  two  more  terms  : 

9.  1  +  2  X  -  3  x2  +  6  x3  -  7  X*  +  10  x5  -  11  x6  +  .... 

10.  1  -  2  X  -  x2  -  7  x3  -  18  x*  -  59  x5  -  181  x^  +  .... 

11.  2  -  11  X  +  15  x2  +  20  x3  -  133  x^  +  231  x^  +  130  x^  +  .... 


THE  DIFFERENTIAL  METHOD 

663.  If  the  first  term  of  a  series  be  subtracted  from  the 
second,  the  second  from  the  third,  and  so  on,  a  series  is  formed 
which  is  called  the  Jlrst  order  of  differences  of  the  given  series. 

The  first  order  of  differences  of  this  new  series  is  called  the 
second  order  of  differences  of  the  given  series ;  and  so  on. 

Thus,  in  the  series 

1,    8,    27,     64,    125,    216,     .••, 
the  successive  orders  of  differences  are  as  follows : 

1st  order,  7,     19,     37,     61,     91,     ••.. 

2d    order,  12,     18,     24,     30,     •••. 

Sd    order,  6,       6,       6, 

4th  order,  0,       0, 


464  ADVANCED  COURSE  IN   ALGEBRA 

The  Differential  Method  is  a  method  for  finding  any  term,  or 
the  sum  of  any  number  of  terms  of  a  series,  by  means  of  its 
successive  orders  of  differences. 

664.    To  find  any  term  of  the  series 

%,       a^i      %,       0^4,       •••,      a„,       Ojn+ly       '"' 

The  successive  orders  of  differences  are  as  follows : 
1st  order,     Og  — tti,  a-g  — as?  cii  —  ds,  '",  «n+i  — ^n;  **'• 
2d  order,     ag  — 2a2  +  a],  (14— 2o3  +  a2,  •••. 
3d  order,     (14  —  3  ag  +  3  ^2  —  ^d  •  •  •  5  etc. 
Denoting  the  first  terms  of  the  1st,  2d,  3d,  •••,  orders  of  dif- 
ferences by  di,  d2,  c?3,  •••,  respectively,  we  have 

di  =  a2  —  (h',  whence,  otg  =  %  +  ^i- 

^2  =  ttg  —  2  a2  +  «! ;  whence, 

ttg  =  —  ai  +  2  tta  +  c?2  =  —  oti  +  2  %  +  2  c?i  +  (?2  =  <^  +  2  c?i  +  (^2- 

dg  =  a4  —  3  ag  +  3  as  —  % ;  whence, 

a4  =  a^  —  3  tta  +  3  ag  -h  c?3  =  tti  -h  3  dj  +  3  ^2  +  c?3 ;  etc. 

It  will  be  observed,  in  the  values  of  as,  a^,  and  a^,  that  the 
coefficients  of  the  terms  are  the  same  as  the  coefficients  of  the 
terms  in  the  expansion  by  the  Binomial  Theorem  of  a  +  a;  to 
the  ^rs^,  second,  and  third  powers,  respectively. 

We  will  now  prove  by  Mathematical  Induction  that  this  law 
holds  for  any  term  of  the  given  series. 

Assume  the  law  to  hold  for  the  nth  term,  a„ ;  then  the  coef- 
ficients of  the  terms  will  be  the  same  as  the  coefficients  of  the 
terms  in  the  expansion  by  the  Binomial  Theorem  of  a  +  ic  to 
the  (n  —  l)th  power ;  that  is, 

,   ,        .,  ,    ,  (n-l)(w-2)  , 
[^ 
^(n-l)(n-2)(n-3)^^^..._         ^^^ 

If  the  law  holds  for  the  wth  term  of  any  series,  it  must  also 
hold  for  the  nth.  term  of  the  first  order  of  differences. 


SUMMATION  OF   SERIES  465 

Or,   a„^,  -  a.  =  d,  -\-(n-l)d,  +  (^-l)(^-2)  ^j^  _^  ...^        ^2) 
Adding  (1)  and  (2),  we  have 
<»„+i  =  ai  +  [(»-l)+l](«.  +  ^[(»-2)  +  2]d2 

This  result  is  in  accordance  with  the  above  law. 

Hence,  if  the  law  holds  for  the  nth  term  of  the  given  series, 
it  holds  for  the  {n  -f  l)th  term  ;  but  we  know  that  it  holds  for 
the  fourth  term,  and  hence  it  holds  for  the  fifth  term;  and 
so  on. 

Therefore,  (1)  holds  for  any  term  of  the  given  series. 

If  the  differences  finally  become  zero,  the  value  of  an  can  be  obtained 
exactly. 

665.  To  find  the  sum  of  the  first  n  terms  of  the  series 

ai,  cfca,  ag,  a^,  a„  ....  (1) 

Let  S  denote  the  sum  of  the  first  n  terms. 
Then  S  is  the  (n  +  l)th  term  of  the  series 

0,  ai,  ai  +  aa,  ai  +  aa  +  agj  ••••  (2) 

The  first  order  of  differences  of  (2)  is  series  (1) ;  whence, 
the  rth  order  of  differences  of  (2)  is  the  same  as  the  (r  —  l)th 
order  of  differences  of  (1). 

Then,  if  di,  do,  ••.,  represent  the  first  terms  of  the  1st,  2d,  •••, 
orders  of  differences  of  (1),  a^,  cZ^,  c?2,  .••,  will  be  the  first  terms 
of  the  1st,  2d,  3d,  •.•,  orders  of  differences  of  (2). 

Putting  (Xi  =  0,  di  =  «!,  (^2  =  c?i?  etc.,  in  (3),  §  664, 

^  =  na,  +  ^%lA)d,  +  "("-|H^^-^)d2+->.         (3) 

666.  Ex.  Find  the  twelfth  term,  and  the  sum  of  the  first 
twelve  terms,  of  the  series  1,  8,  27,  64,  125,  .... 


466  ADVANCED   COURSE   IN   ALGEBRA 

Here,  n  =  12,  a^  =  1. 

Also,  di  =  7,  ^2=12,  (^3  =  6,  and  f^4  =  0  (§663). 

Substituting  in  (1),  §  664,  the  twelfth  term 

=  1  4- 11 .  7  +  llli^  .  12  +  ^^  '\^'^  .  6  =  1728. 

Substituting  in  (3),  §  665,  the  sum  of  the  first  twelve  terms 
=  12  +  iilll .  7  +  l^^iill^ .  12  + 1241:^ .  6  =  6084. 

667.   Piles  of  Shot. 

Ex.  If  shot  be  piled  in  the  shape  of  a  pyramid  with  a  tri- 
angular base,  each  side  of  which  exhibits  9  shot,  find  the  num- 
ber in  the  pile. 

The  number  of  shot  in  the  first  five  courses  are  1,  3,  6,  10, 
and  15,  respectively ;  we  have  then  to  find  the  sum  of  the  first 
nine  terms  of  the 'series  1,  3,  6,  10,  15,  •••. 

The  successive  orders  of  differences  are  as  follows : 

1st  order,  2,     3,     4,     5,     •••. 

2d  order,  1,     1,     1,     •••. 

3d  order,  0,     0,     .... 

Putting  ?i  =  9,  «!  =  1,  di  =  2,  c\  =  1  in  (3),  §  665, 

EXERCISE  no 

1.  Pind  the  first  term  of  the  sixth  order  of  differences  of  the  series 
3,  5,  11,  27,  67,  159,  375,  .... 

2.  Find  the  15th  term,  and  the  sum  of  the  first  15  terms,  of  the  series 

1,  9,  21,  37,  57,  .... 

3.  Find  the  14th  term,  and  the  sum  of  the  first  14  terms,  of  the  series 
5,  14,  15,  8,  -7,  .... 

4.  Find  the  sum  of  the  first  n  multiples  of  3. 

6.   Find  the  nth  term,  and  the  sum  of  the  first  n  terms,  of  the  series 

2,  -  1,  1,  8,  20,  .... 


SUMMATION  OF   SERIES  467 

6.  If  shot  be  piled  in  the  shape  of  a  pyramid  with  a  square  base,  each 
side  of  which  exhibits  25  shot,  find  the  number  in  the  pile. 

7. 1'ind  the  13th  term,  and  the  sum  of  the  first  13  terms,  of  the  series  1, 
3,9,25,  57,111,  -. 

8.  Find  the  10th  term,  and  the  sum  of  the  first  10  terms,  of  the  series  4, 
-2,10,  4,  -56,-206,  ....' 

9.  Find  the  sum  of  the  squares  of  the  first  n  multiples  of  2. 

10.  Find  the  n\h  term,  and  the  sum  of  the  first  n  terms,  of  the  series  1, 
-3,-13,-  17,  -3,  41,  .... 

11.  Find  the  number  of  shot  in  a  pile  of  9  courses,  with  a  rectangular 
base,  if  the  number  of  shot  in  the  longest  side  of  the  base  is  24. 

12.  Find  the  number  of  shot  in  a  truncated  pile  of  10  courses,  with  a 
square  base,  if  the  number  of  shot  in  each  side  of  the  lower  base  is  16. 

13.  Find  the  number  of  shot  in  a  truncated  pile  of  8  courses,  with  a 
rectangular  base,  if  the  number  of  shot  in  the  length  and  breadth  of  the 
base  are  20  and  14,  respectively. 

14.  Find  the  12th  term,  and  the  sum  of  the  first  12  terms,  of  the  series 
1,  13,  49,  139,  333,  701,  1333,  .... 

15.  Find  the  9th  term,  and 'the  sum  of  the  first  9  terms,  of  the  series 
20,  4,  -  36,  -  132,  -  356,  -  820,  -  1676,  .... 

16.  Find  the  sum  of  the  fourth  powers  of  the  first  n  natural  numbers. 

17.  Find  the  number  of  shot  in  a  pile  with  a  rectangular  base,  if  the 
number  of  .  shot  in  the  length  and  breadth  of  the  base  are  m  and  w, 
respectively. 

18.  Find  the  number  of  shot  in  a  truncated  pile  of  n  courses,  with  a 
triangular  base,  if  the  number  of  shot  in  each  side  of  the  lower  base  is  m. 

INTERPOLATION 

668.  Interpolation  is  the  process  of  introducing  between  the 
terms  of  a  series  other  terms  conforming  to  the  law  of  the 
series. 

Its  usual  application  is  in  finding  intermediate  numbers 
between  those  given  in  Mathematical  Tables. 

The  operation  is  effected  by  giving  fractional  values  to  n  in  (1), 
§  664. 

The  method  of  Interpolation  rests  on  the  assumption  that  a 
formula  which  has  been  proved  for  an  integral  value  of  n,  holds 
also  when  n  is  fractional. 


468       ADVANCED  COURSE  IN  ALGEBRA 

Ex.      Given      V5  =  2.2361,      V6  =  2.4495,      V7  =  2.6458, 
V8- 2.8284,  ...;  find  V6.3. 
In  this  case  the  successive  orders  of  differences  are : 
.2134,     .1963,     .1826,     .... 
-.0171,   -.0137,  .... 
.0034,  .... 

Whence,  d^  =  .2134,   d^  =  -  .0171,  d^  =  .0034,  .... 

Now,  the  required  term  is  distant  1.3  intervals  from  V5. 

Substituting  n  =  2.3  in  (1),  §  664,  we  have,  approximately, 

V6:3  =  2.2361  +  1.3  x  .2134  +  ^f  ^  f  (-  .0171) 

1x2^ 

.  1.3  X  •3x-.7^_()Q3^ 


1x2x3 
=  2.2361  +  .2774  -  .0033  -  .0002  =  2.5100. 

EXERCISE    III 

1.  Given  log 26  =  1.4150,  log 27  =  1.4314,  log28  =  l. 4472,  log  29=1.4624, 
...  ;  find  log 26. 7. 

2.  Given  v^91  =  4.49794,  v/92  =4.51436,  v^93=4.53066,  \/94=4.54684, 
...;  find  v'92.5. 

3.  The  reciprocal  of  35  is  .02857  ;  of  36,  .02778  ;  of  37,  .02703  ;  of  38, 
.02032 ;  etc.     Find  the  reciprocal  of  36.28. 

4.  Given  log  124  =  2.09342,  log  125  =  2.09691,  log  126  =  2.10037, 
log  127  =  2.10380,  ...  ;  find  log  125.36. 

6.  Given  21^  =  9261,  22^  =  10648,  23^  =  12167,  24^  =  13824,  and 
253  =  15625  ;  find  the  cube  of  2H. 

6.  Given  log  61  =  1.78533,  log  62  =  1.79239,  log  63  =  1.79934, 
log  64  =  1.80618,  ...  ;  find  log 63.527. 

SUMMATION  OF    SERIES    BY  SEPARATION  INTO  PARTIAL 
FRACTIONS 

318.  1.  Find  the  sum  of  the  first  n  terms  of  the  infinite 
series  i  i  i 

2.3     3.4     4.5         ' 
and  determine  whether  the  series  is  convergent  or  divergent. 


SUMMATION   OF   SERIES  469 

The  nth.  term  of  the  above  series  is — ; • 

(n  +  l)(n  +  2) 

Separating  into  partial  fractions  by  the  method  of  §  567,  we 
have  ^  11 


(n  +  1)  (n-\-2)      71  +  1      n-\-2 
Then  the  sum  of  the  first  n  terms  of  the  given  series  is 

(|-|)-(l-i)K14)--(.-Ti-.-i2) 

11  n 


2     n-i-2     2(n-^2) 

11  1 

Since approaches  the  limit  -,  when  7i  is  indefinitely 

2     n  -\-2  2 

increased,  the  series  is  convergent  (§  543). 

2.   Find  the  sum  of  the  first  n  terms  of  the  infinite  series 


1-3     2-4     3-5  ■       ' 
and  determine  whether  the  series  is  convergent  or  divergent. 

The  nth  term  of  the  series  is  — — ,  which,  by  the  method 

n{n  +  2) 

of  §  567,  equals  A -1^^. 
2  n     2{n  +  2) 

Then  the  sum  of  the  first  n  terms  is 

\2     6;      V4     ^)      \^      lOy^         [_2n      2(n  +  2)J 

All  the  terms  cancel  except  the  first  two  positive,  and  the 
last  two  negative. 

Thus  the  sum  of  the  first  n  terms  is 

1,11  1  3  1  1 


2     4     2(n  +  l)      2(n  +  2)'        4     2(n  +  l)      2(7i  +  2) 

3 

The   latter  expression  approaches  the  limit   -   when  n   is 

indefinitely  increased. 

Hence,  the  series  is  convergent. 


470  ADVANCED   COURSE   IN   ALGEBRA 


EXERCISE  112 

In  each  of  the  following,  find  the  sum  of  the  first  n  terms,  and  deter- 
mine whether  the  series  is  convergent  or  divergent : 

2. 53.  10     4. 15  1-42.53.6 

12  .  22  ^  22  .  32        32  .  42 

In  each  of  the  following,  find  the  sum  of  the  first  n  terms,  and  deter- 
mine for  what  values  of  x  the  series  is  convergent  : 

-      1    +,  ..A  ...  +  .    J.    „,  +  .■■■ 


a;(x  +  l)      (x+l)(a;  +  2)      (x  +  2)(a:  +  3) 


(l+x)(l  +  2x)      (l  +  2x)(l  +  3x)      (l  +  8x)(l  +  4x) 
8.   Find  the  sum  of  the  first  n  terms  of  : 

1        +,    .,,  1  .,    „.  +  ■■■■ 


«(x  +  l)(a:  +  2)      (x  +  l)(a;  +  2)(a;  +  3) 


THEORY   OF   NUMBERS     •  471 


XXXV.    THEORY  OF  NUMBERS 

670.  In  the  present  chapter,  the  word  number  will  signify  a 
positive  integer;  and  every  letter  will  be  understood  as  repre- 
senting a  positive  integer. 

One  number  is  said  to  divide  another  when  it  is  contained  in 
it  without  a  remainder ;  in  this  case,  the  second  number  is  said 
to  be  a  multiple  of  the  first. 

A  prime  number  is  a  number  which  cannot  be  divided,  with- 
out a  remainder,  by  any  number  except  itself  and  unity. 

One  number  is  said  to  be  prime  to  another  wnen  there  is  no 
number,  except  unity,  which  will  divide  each  of  them  without 
a  remainder. 

671.  If  a  number  divides  the  product  of  two  others^  and  is 
prime  to  one  of  them,  it  must  divide  the  other. 

Let  the  number  n  divide  the  product  ab,  and  be  prime  to  a. 

Since  n  divides  ab,  the  prime  factors  of  ab  must  be  the  same 
as  those  of  n,  with  certain  additional  prime  factors. 

But  since  n  is  prime  to  a,  n  and  a  have  no  common  factor 
except  unity. 

Then,  the  prime  factors  of  b  must  be  either  the  same  as  those 
of  n,  or  with  certain  additional  prime  factors ;  and  n  divides  b. 

672.  If  a  prime  number  divides  the  product  of  any  number  of 
factors,  it  must  divide  some  f alitor  of  the  product. 

Letp  be  a  prime  number  which  divides  the  product  abc  •••. 

Then,  the  prime  factors  of  abc  •••  must  hep,  with  certain 
additional  prime  factors. 

Then,  some  one  of  the  numbers  a,  b,  c,  '•-,  must  have  p  as  a 
prime  factor,  and  therefore  p  divides  some  factor  of  the  product. 

673.  It  follows  from  §  672  that 

If  a  prime  number  divides  a  positive  integral  power  of  a  num- 
ber a,  it  must  divide  a. 


472  ADVANCED  COURSE  IN   ALGEBRA 

674.  If  the  numerator  and  denominator  of  a  fraction  are 
prime  to  each  other,  the  fraction  is  in  its  lowest  terms. 

Let  a  be  prime  to  b. 

If  possible,  let  -  =  — ,  where  b'  is  <  b. 

^  ab' 

Multiplying  both  members  by  b',  a'  =  — • 

Whence,  since  a'  is  an  integer,  b  divides  ab'. 

But  a  is  prime  to  b,  and  hence  b  divides  b'  (§  671). 

But  this  is  impossible  if  b'  is  <  b. 

Hence,  b'  cannot  be  <  b,  and  -  is  in  its  lowest  terms. 

b 

675.  It  was  proved,  in  §  674,  that  b  divides  b'. 
Whence,  b'  =  mb,  where  m  is  an  integer. 

rnr,  ,     ab'      amb 

Then,  a' =  —  =  -—~  =  am. 

b          b 

Hence,  if  a  is  prime  to  b,  and  ~  =  — ,  a'  and  b'  are  equimulti- 
ples of  a  and  b. 

676.  A  number  can  be  resolved  into  prime  factors  in  only  one 
way. 

Let  w  be  a  composite  number ;  and  suppose  n  —  abc  •••,  where 
a,  b,  c,  •••,  are  prime  numbers. 

Suppose  also,  if  possible,  that  n  =  a'b'c'  •••,  where  a',  &',•  c',  •••, 
is  a  different  set  of  prime  numbers. 

We  have,  a6c  •••  =  a'6'c' ••-.  (1) 

Then,  a  divides  a'b'c'  •",  and  therefore  divides  some  one  of 
the  numbers  a',  b',  c',  •••  (§  672). 

Then,  since  a,b,c,  -",  and  a',  b',  c',  •••,  are  prime  numbers,  a 
must  equal  some  one  of  the  numbers  a',  b',  c',  •••.  "-- 

Suppose,  then,  a  =  a'  -,  dividing  the  members  of  (1)  by  a  and 
a',  respectively,  ^^...  ^^/c'-..; 

and,  as  before,  b  equals  one  of  the  numbers  b',  c',  •••. 

Proceeding  in  this  way,  we  can  prove  each  of  the  numbers 
a,  bf  c,  •••,  equal  to  one  of  the  numbers  a',  b',  c',  •••. 


THEORY   OF  NUMBERS  478 

677.   To  find  the  highest  power  of  2  which  is  contained  in  [14. 

Of  the  factors  1,  2,  3,  ••-,  14,  the  numbers  2,  2  x  2,  3  x  2,  ••., 
7x2  contain  2  as  a  factor  at  least  once. 

The  numbers  2^,  2  x  2^,  3  x  2^  contain  2^  as  a  factor  at  least 
once. 

The  number  2^  contains  2^  as  a  factor  once. 

Then,  the  highest  power  of  2  in  [14  is  evidently 

7+3  +  1,  or  11. 

We  will  now  consider  the  general  case. 

To  find  the  highest  power  of  a,  where  a  is  any  prime  number, 
which  is  contaiyied  in  \n. 

Of  the  factors  1,  2,  3,  •••,  n,  the  numbers  a,  2  a,  3  a,  •••,  con- 
tain a  as  a  factor  at  least  once. 

Let  the  last  term  of  this  series  be  pa. 
Then,  pa  either  equals  w,  or  is  <  n ;  whence, 

n  ^n 

p  =  -,  ov  p<  — 

a  a 

Therefore,  p  is  the  greatest  integer  in  — 

Again,  the  numbers  a^,  2a^,  Sa^,  •••,  contain  a^  as  a  factor  at 
least  once ;  let  the  last  term  of  this  series  be  qa\ 

Then,  as  before,         q  =  —^,  or  q<—' 
a^  a- 

Whence,  q  is  the  greatest  integer  in  — • 

ar 

Continuing  in  this  way,  the  highest  power  of  a  in  \n  is 

i>  +  g  +  r+..-, 

where  p  is  the  greatest  integer  in  -,  q  in  — ,  r  in  — .  etc, 

a  (T  (T 

We  will  now  solve  the  example  of  §  677  by  this  method. 

In  this  case,  a  =  2,  w  =  14  ;  then,  t*  =  7,  ^  =  I,  -^  t=  I 

7  7 

The  greatest  integer  in  7  is  7  ;  in  -,  is  3  ;  in  -,  is  1. 

2  4 

Then,  the  highest  power  of  2  in  [14  is  7  +  3  +  1,  or  11. 


474  ADVANCED   COURSE   IN   ALGEBRA 

678.    The  product  of  any  n  consecutive  integers  is  divisible  by  [n. 

Let  the  integers  be  m  +  1,  m  -\-  2,  •••,  m  -\-  n. 
Multiplying  both  numerator  and  denominator  by  the  prod- 
uct of  the  natural  numbers  from  1  to  m,  inclusive,  the  fraction     i 

(m  4-  1)  (m  +  2) .-.  (m  4-  n)  ^  1^^  +  ^ 

\n  \m\n 

If  a  is  any  prime  number,  the  exponent  of  a  in  \m  -^  7i  is 

7n  -\-  11 
p  +  g  +  r-f  •••,  where  p  is   the  greatest  integer  in    , 

q  in  — =^,  r  in        \     ,  etc.  (§  677). 

a'^  a^ 

The  exponent  of  a  in  [m  is  pi  +  ^i  +  r^  H ,  where  p^  is  the 

,,.,  .     m       m       ■     m      , 

greatest  integer  m  — ,  g^  m  — ,  r^  m  — ,  etc. 
a  a^  or 


The  exponent  of  a  in  |^  is  pg  +  Q'2  +  ^2  +  •*•?  where  p2  is  the 

greatest  integer  m  -,  gg  m  —  ?  ^2  m  —  j  ©tc. 
tt  a^  a^ 

Then,  the  exponent  of  a  in  |m[9i  is  {pi  +p^  +  fe  +  ^'2)  +  •••• 
Now,  the  greatest  integer  in  is  either  the  sum  of  the 

greatest  integers  in  —  and  -,  or  else  it  exceeds  this  sum 
by  unity. 

That  is,  p  equals  Pi  -{-P2,  or  Pi-\-p2-\- 1. 

Similarly,  q  equals  gi  +  gg,  or  gi  +  g2  +  1 ;  etc. 

Then,  p  ^  q  ^  r  -] is  not  <  (pi  +  q^)  +  {P2  +  ^2)  +  ••• ; 

that  is,  the  exponent  of  a  in  \m  +  n  is  not  <  its  exponent  in 
\m\n. 

^,         Im  +  n 

Then,  must  be  an  integer. 

\m\n 

Whence,  (m  + 1)  (m  4-  2)  •  •  •  (m  +  n)  is  divisible  by  \n. 

679.  If  n  is  a  prime  number,  the  coefficieiit  of  any  term,  except 
the  first  and  last,  in  the  expansion  by  the  Binomial  Theorem  of 
{a  +  xy,  is  divisible  by  n. 


THEORY  OF   NUMBERS  475 

By  §  287,  the  coefficient  of  the  (r  +  l)th  term  of  (a  +  a;)"  is 

\r 
By  §  678,  n  (n  —  1)  '•'  (n  —  r  -{- 1)  is  divisible  by  |r. 
If  r  has  any  value  from  1  to  n  —  1,  none  of  the  factors  of  [r 
,  can  divide  n]  for,  by  hypothesis,  n  is  a  prime  number. 
Then,  \r  must  divide  (w  —  1)  •••  (n  —  ?-  +  1). 
Therefore,  expression  (1)  is  an  integer,  and  divisible  by  n. 

680.  Fermat's  Theorem. 

If  71  is  a  prime  number,  and  m  is  prime  to  n,  m"~^  —  1  is  a 
multiple  of  71. 

Let  a,  b,  c,  '•',  k  be  any  m  numbers. 
Expanding  by  the  Binomial  Theorem,  we  have 
(a  +  6  +  c  +  ...  +  A;)"  =  [a  +  (&  +  c  +  -  +  k)Y 
=  a"  +  na"  '  (?>  +  c  4-  •  •  •  +  fc)  +  ''^(^-^)  a«-2  (6  4-  c  +  . . .  +  A;)2 
+  -+(6  +  c  +  ...+A;)".        *-  .  (1) 

By  §  679,  each  coefficient  n,  ^^^~    ^^  -..,  is  divisible  by  n. 

Then,  every  term  in  the  expansion  (1),  which  is  not  a  power 
of  a,  b,  c,  ...,  k,  is  a  multiple  of  n;  and  (1)  can  be  written 

(a  +  &  +  c 4-  •••  +  A;)"  =  (a»  +  &"  +  c"  +  ♦••  +  A;")  +_pn, 
where  p  is  an  integer. 

Putting  a  =  b  =  c=  •"  =k=l,  we  have 

m'^  =  m  -i-pn,  or  m"  —  m  =pn,  or  m  (m"~^  —  1)  =pn. 
Since  m  is  prime  to  n,  it  cannot  be  a  multiple  of  n. 
Then,  m"7^  —  1  must  be  a  multiple  of  n. 

681.  The  result  m^'  —  m^pn,  of  §  680,  holds  if  m  is  not 
prime  to  n. 

That  is,  m"  —  m  is  a  multiple^f  n  when  7i  is  prime,  whether 
m  is  prime  to  n  or  not. 


476  ADVANCED  COURSE  IN  ALGEBRA 

EXAMPLES 

682.  1.  If  two  consecutive  numbers  are  not  multiples  of  3, 
their  sum  is  a  multiple  of  3. 

The  numbers  must  be  in  the  forms  3  m  + 1  and  3  m  +  2. 

Then  their  sum  is  6  m  +  3,  which  is  a  multiple  of  3. 

It  follows  from  the  above  that  any  number  of  the  form 

n(n  +  1)  [w  +  (w  +  1)],  or  n(n  +  1)  (2  n  +  1), 
is  divisible  by  6. 

For  either  n  or  n  +  1  must  be  even  ;  and  if  neither  n  nor  w  +  1  is  a 
multiple  of  3,  their  sum  is  a  multiple  of  3. 

2.  Every  perfect  square  is  in  the  form  5  n,  ov  5n  ±  1. 

Tor  any  number  is  in  the  form  5  m,  or  5  m  ±  a,  where  a  may 
be  1  or  2. 

(5  my  is  in  the  form  5  n. 

Again,  {5  m  ±  ay  equals  a  multiple  of  5,  plus  al 

But  a^  is  either  1  or  4 ;  and  4  =  5  —  1. 

Hence,  (5  m  ±  ay  differs  by  1  from  a  multiple  of  5,  and  is 
therefore  in  the  form  5n-\-l  or  5n  —  l. 

3.  If  n  is  even,  and  not  a  multiple  of  3,  n^  -\-2  is  divisible 
by  6. 

For  n  must  be  in  the  form  3  m  ±  1,  where  m  is  odd. 
Then,  n^ -\-2  =  9  m^  ±6  m-\-l  +  2,  which  is  divisible  by  3. 
Again,  n^-\-2  is  even;  hence,  it  is  divisible  by  6. 

4.  Every  even  power  of  an  odd  number  is  in  the  form 
-8  n  + 1. 

Now,  any  odd  number  is  in  one  of  the  forms 
8  m  ±  1,  or  8  m  ±  3. 

If  this  be  raised  to  any  even  power,  the  result  will  be  a 
multiple  of  8  plus  1,  or  a  multiple  of  8  plus  an  even  power 
of  3. 

Now  any  even  power  of  3  is  in  the  form  of  a  multiple  of 
8,  plus  1. 

Hence,  any  even  power  of  an  odd  number  is  in  the  form 
8n  +  l. 


THEORY  OF  NUMBERS  477 

5.  Prove  that  n'  —  n  is  divisible  by  42. 

By  §  681,  since  7  is  prime,  n''  —  n  is  divisible  by  7. 

Again,  nJ  —  n  =  n  {n^  —  l)  =  n  (n^  —  1)  (71'^  -f  n^  + 1) 

=  (n  -  1)  n  (n  + 1)  (n'^  +  ri^  + 1). 

By  §  678,  (n  -1)  n  (n  +  1)  is  divisible  by  [3,  or  6. 
Hence,  -nJ  —  n  is  divisible  by  7  and  6,  or  by  42. 

6.  Prove  that  the  fourth  power  of  any  number  is  in  the 
form  5  71  or  5  ?i  +  1- 

If  m  is  prime  to  5,  m''  —  1  is  a  niultiple  of  5  (§  680). 

That  is,  m*  is  a  multiple  of  5,  increased  by  1 ;  and  is  there- 
fore in  the  form  5n  -\-l. 

If  m  is  not  prime  to  5,  it  contains  5  as  a  factor;  and  m^  is 
in  the  form  5  n. 

EXERCISE  113 

1.  If  n  is  odd,  and  greater  than  1,   (n  —  1)  n  (n  +  1)  is  divisible 
by  24. 

2.  Prove  that  n^  —  n  is  divisible  by  30,  if  n  is  greater  than  1. 

3.  Prove  that  w  (w  +  1)  (n  +  5)  is  a  multiple  of  6. 

4.  Find  the  highest  power  of  2  in  |^. 

5.  Find  the  highest  power  of  3  in  [80. 

6.  Prove  that  every  perfect  square  is  in  the  form  3  n,  or  3  n  +  1- 

7.  Prove  that  every  perfect  sixth  power  is  in  the  form  7  n,  or  7  w  +  1. 

8.  Prove  that,  if  w  is  odd,  and  not  a  multiple  of  3,  n^  +  5  is  divisible 
byC. 

9.  Prove  that,  if  n  is  odd,  (71^  +  3)  (w^  +  7)  is  divisible  by  32. 

10.  Prove  that  any  odd  power  of  7  is  in  the  form  8  n  —  1. 

11.  Prove  that  w^  —  5  n^  +  4  »  is  divisible  by  120,  if  n  is  greater  than  2. 

12.  Prove  that,  if  w  is  odd,  and  greater  than  1,  n^  —  n  is  divisible 
by  240. 

13.  Prove  that  every  perfect  cube  is  in  the  form  7  w,  or  7  n  i  1. 

14.  Prove  that  every  perfect  cube  is  in  the  form  9  n,  or  9  w  ±  1. 


478 


ADVANCED   COUKSE   IN   ALGEBRA 


XXXVI.     DETERMINANTS 


IS 


683.   The  solution  of  the  equations 
a^x  +  bi,y  =  Ci, 
a^x  +  b^  =  C2, 
hfi  —  61C2         CoO-i  —  Cia2 


x  = 


y 


a-J)2  —  <^2^1  <^1^2  —  ^2^1 

The  common  denominator  may  be  written  in  the  form 


(1) 


This  is  understood  as  signifying  the  product  of  the  upper 
left-hand  and  lower  right-hand  numbers,  minus  the  product  of 
the  lower  left-hand  and  upper  right-hand. 

The  expression  (1)  is  called  a  Determinant  of  the  Second 
Orde>. 

The  numerators  of  the  above  fractions  can  also  be  expressed  as  deter- 
minants ;  thus,  ■  ^ 


1,                     X.                       ^ll 

02CI  -  01C2  =   ^ 

02, 

&2 

,  and  C2ai 

^1'     ^^ 
-cia2=   ^      ^    • 
a2,     C2 

684.   The  solution  of  the  equations 

r  a,x  +  b^tj  +  c,z  =  di, 

a^  +  ^22/  +  G^  =  d2, 

I    agOJ    +    &3?/    +    C32;    =    dg, 

.„          ^     d,h2C,  -  d,b,C2  +  ^2?>3Ci  -  ^2&iC3  4-  dsb,C2  -  d,b2Ci 

C^A^S  —  CtAC'I  +  «2&3Cl  —  «2&lC3  +  «3&lC2  "  ^3 Vl 

with  results  of  similar  form  for  y  and  2;. 

The  denominator  of  (1)  may  be  written  in  the  form 

«!,       5i,       Ci 

a2,     b2,     C2 

• 

% 

&3,      c? 

(1) 


(2) 


DETERMINANTS 


479 


This  is  understood  as  signifying  the  sura  of  the  products  of 
the  numbers  connected  by  lines  parallel  to  a  line  joining  the 
upper  left-hand  corner  to  the  lower  right-hand,  in  the  follow- 
ing diagram,  minus  the  sum  of  the  products  of  the  numbers 
connected  by  lines  parallel  .to  a  line  joining  the  lower  left-hand 
corner  to  the  upper  right-hand. 


Y  1 


The  expression  (2)  is  called  a  Determinant  of  the  Third  Order. 

The  numerator  of  the  value  of  x  can  also  be  expressed  as  a  determinant, 


as  follows : 


as  may  be  verified  by  expanding  it  by  the  above  rule. 

685.  The  numbers  in  the  first,  second,  etc.,  horizontal  lines, 
of  a  determinant,  are  said  to  be  in  the  Jirst,  second,  etc.,  7'ows, 
respectively ;  and  the  numbers  in  the  first,  second,  etc.,  vertical 
columns,  in  th^  Jirst,  second,  etc.,  columns. 

The  numbers  constituting  the  determinant  are  called  its 
eletneyits,  and  the  products  in  the  expanded  form  its  terms. 

Thus,  in  the  determinant  (2),  of  §  684,  the  elements  are 
«i>  <^2j  %)  ^tc,  and  the  terms  a^h^Cz,  —  a-J^^Cz,  etc. 


EXERCISE  114 


Evaluate  the  following  determinants  : 

"x-f-y,    x-y\  2    \a-h,     -2a 

x  —  y,    x-{-  y\'  *    I      2  6,     a  —  h 


1. 


1,     5, 

2 

3. 

4,     7, 

3  . 

9,     8, 

6 

480 


ADVANCED  COURSE  IN  ALGEBRA 


6,    4 

,     7 

2,     - 

3,         11 

9,     0,     8 

. 

5. 

-  2,          4,          5 

5,     3,     2 

3,     -1,     -4 

a,     b, 

c 

a^  +  y, 

7. 

&,     a, 

(Z 

. 

8. 

X, 

c. 

d, 

a 

2/, 

1, 

X, 

a 

1,    y, 

& 

1,   ^, 

c 

;2 

X 

. 

0  +  x 

^> 

Verify  the  following  by  expanding  the  determinants  : 

as, 

11.      aa, 

as, 

686.  If,  in  any  permutation  of  the  numbers  1,  2,  3,  •••,  w,  a 
greater  number  precedes  a  less,  there  is  said  to  be  an  inversion. 

Thus,  in  the  case  of  five  numbers,  the  permutation  4,  3,  1,  5,  2  has  six 
inversions ;  4  before  1,  3  before  1,  4  before  2,  3  before  2,  5  before  2,  and 
4  before  3. 

Consider,  now,  the  n^  elements 


6l,     Ci 

«s,   63,    as 

wai, 

61,    Ci 

ai,   mbu   ci 

62,  C2 

— 

C2,   &2,   a2 

10. 

?)ia2,    62,   C2 

= 

a2,   W62,   C2 

63,    Cs 

ci,    61,   ai 

mas,   6s,   Cs 

as,   W63,   Cs 

61,  c 

62,  C 

63,  c 

I 
2 

3 

= 

^     62,     C2 
:  ai    , 

63,     Cs 

-hi 

ao,     C2 
as,     Cs 

as. 

62 
63  ' 

^2,l> 


''2,2? 


^-2,3? 


ag 


(1) 


XI?       "'n,2j       "■m,3?        •"?        ^»i 

The  notation  in  regard  to  suffixes,  in  (1),  is  that  the  first 
suffix  denotes  the  row,  and  the  second  the  column,  in  which 
the  element  is  situated. 

Thus,  aj,^ ,.  is  the  element  in  the  /cth  row  and  ?'th  column. 

Let  all  possible  products  of  the  elements  taken  n  at  a  time  be 
formed,  subject  to  the  restriction  that  each  product  shall  con- 
tain one  and  only  one  element  from  each  row,  and  one  and  only 
one  from  each  column,  and  write  them  so  that  the  first  suffixes 
shall  be  in  the  order  1,  2,  3,  •••,  n. 

This  is  equivalent  to  writing  all  the  permutations  of  the  numbers  1,  2, 
•  ",  w  in  the  second  suffixes. 

Make  each  product  +  or  —  according  as  the  number  of  inver- 
sions in  the  second  suffixes  is  even  or  odd. 


DETERMINANTS  481 

The  expression  (1)  is  called  a  Determinant  of  the  nth  Order. 

The  number  of  terms  in  the  expanded  form  of  a  determinant  of  the  nth 
order  is  |22^(§  625). 

The  elements  lying  in  the  diagonal  joining  the  upper  left- 
hand  to  the  lower  right-hand  corner,  are  said  to  be  in  the 
principal  diagonal;  the  term  whose  factors  are  the  elements  in 
the  principal  diagonal  is  always  positive. 

The  determinant  (1)  of  §  686  is  sometimes  expressed  by  writing  only 
the  elements  in  the  principal  diagonal ;  thus, 

I  «i,  1)    «2, 2,    •••?    «„,„!• 
When  in  the  form  (1)  of  §  686,  the  determinant  is  said  to  be  in  the 
Square  Form. 

687.  It  may  be  shown  that  the  definition  of  §  Q>^Q  agrees 
with  that  of  §  684. 

For  consider  the  determinant  ,,   ,.^ 

<^l,  1)  <^l,  2J  <%,  3 
^2,  Ij  ^2,  2j  ^2, 3 
<^3, 1^     <^3, 2J     <^3, 3 

The  products  of  the  elements  taken  3  at  a  time,  subject  to 
the  restriction  that  each  product  shall  contain  one  and  only 
one  element  from  each  row,  and  one  and  only  one  from  each 
column,  the  first  suffixes  being  written  in  the  order  1,  2,  3,  are 

^1, 1  ^2,  2  ^3,  3)     <^1, 1  ^2,  3  ^3,  2j     %,  2  ^2, 1  ^3,  3?     ^1,  2  <^2, 3  ^3, 1?     <^1,  3  <^2, 1  <^3,  2? 

and  Oi  3  a2, 2  <^3, 1- 

In  the  first  of  these  there  are  no  inversions  in  the  second 
suffixes ;  in  the  second  there  is  one,  3  before  2 ;  in  the  third 
there  is  one;  in  the  fourth,  two;  in  the  fifth,  two;  in  the 
sixth,  three. 

Then  by  the  rule  of  §  686,  the  first,  fourth,  and  fifth  products 
are  positive,  and  the  second,  third,  and  sixth  are  negative ;  and 
the  expanded  form  is 

^,1^2,2^3,3  —  ^,1  ^2,3^3.2  —  ^1.2<^2,1%,3  ~f"  %,2<^2,3^3,1 
~r  C^l.  3  (^2, 1  %,2  —  ^1.  3  ^2,  2  ^Z,  1) 

which  agrees  with  §  684. 


482  ADVANCED   COURSE  IN  ALGEBRA 

688.  If,  in  any  permutation  of  the  letters  1,  2,  3,  •••,  n,  two 
numbers  he  interchanged,  the  number  of  inversions  is  iyicreased 
or  diminished  by  an  odd  number. 

If  two  consecutive  numbers  be  interchanged,  the  number  of 
inversions  is  increased  or  diminished  by  1. 

Thus,  if  •••  aZcmft  •••  is  a  permutation  of  1,  2,  3,  •••,  n,  the 
number  of  inversions  in  •••  ahmb  •••  differs  by  1  from  the  number 
vc\.  •"  amkb  •"•,  for  the  number  of  inversions  so  far  as  the 
numbers  preceding  k,  and  following  m,  is  concerned  is  not 
affected  by  the  interchange. 

Now  let  •••  akbc  •••  efmg  •••  be  a  permutation  of  the  numbers 
1,  2,  3,  •••,  71,  having  |7  numbers  between  h  and  m. 

By  interchanging  m  with  /,  then  m  with  e,  and  so  on  in  suc- 
cession with  each  of  the p-\-l  numbers  to  the  left  of  m,,  m  may 
be  brought  to  the  left  of  k. 

Then,  by  interchanging  k  with  b,  then  k  with  c,  and  so  on  in 
succession  with  each  of  the  p  numbers  to  the  right  of  k,  k  may 
be  brought  to  the  right  of/. 

Thus,  k  and  m  may  be  interchanged  by  p  +  1  +jp,  ov  2p-\-l 
interchanges  of  consecutive  numbers  ;  that  is,  by  an  odd  number 
of  interchanges  of  consecutive  numbers. 

It  follows  from  this  that,  if  k  and  m  be  interchanged,  the 
number  of  inversions  is  increased  or  diminished  by  an  odd 
number. 

689.  Any  permutation  of  the  numbers  1,  2,  3,  •••,  n  can  be 
obtained  from  any  other  by  repeated  interchanges  of  two 
numbers. 

If  we  commence  in  this  way  with  the  permutation  1,  2,  3, 
•••,  n  itself,  after  one  such  interchange  there  will  be  an  odd 
number  of  inversions  in  the  resulting  permutation  (§  688). 

If  in  the  latter  we  interchange  two  numbers,  we  shall  have 
a  permutation  with  an  even  number  of  inversions. 

Proceeding  in  this  way,  since  the  whole  number  of  permuta- 
tions is  an  even  number,  we  shall  find  that  the  number  of  per- 
mutations with  an  odd  number  of  inversions  equals  the  number 
with  an  even  number  of  inversions. 


DETERMINANTS  483 

It  follows  from  the  above  that,  in  the  expanded  form  of  the 
determinant  (1),  of  §  686,  the  number  of  positive  terms  equals 
the  number  of  negative  terms. 

690.  The  expanded  form  of  the  determinant  (1),  §  686,  may 
also  be  obtained  by  writing  the  seco7id  suffixes  in  the  order 
1,  2,  3,  ••♦,  n,  and  making  each  product  +  or  —  according  as  the 
number  of  inversions  in  the  Jirst  suffixes  is  even  or  odd. 

For  let  the  absolute  value  of  any  term,  obtained  by  the  rule 

of  §686,  be  ai,pa2,.-an,.;  (1) 

where  p,  g-,  ••♦,  r  is  a  permutation  of  1,  2,  •••,  n. 
This  is  obtained  from  the  first  term 

by  changing  second  suffixes,  1  to  p,  2  to  g,  •••,71  to  r.  . 

Since  p,  q,"',r  is  a  permutation  of  1,  2,  •••,?i,  (2)  may  be 
written  ^      ^     ...  ^     . 

and  (1)  may  be  obtained  from  this  by  changing  first  suffixes, 
p  to  1,  g  to  2,  •••,  r  to  n. 

In  these  two  methods,  there  is  the  same  number  of  inter- 
changes of  two  suffixes,  and  the  term  (1)  will  therefore  have 
the  same  sign. 

PROPERTIES  OF  DETERMINANTS 

691.  A  determinant  is  not  altered  in  value  if  its  rows  are 
changed  to  columns,  arid  its  columns  to  rows. 

The  truth  of  the  theorem  for  a  determinant  of  the  third  order  may 
easily  be  seen  by  expanding  the  determinants 


«1, 

&i, 

Cl 

ai, 

a2, 

as 

a2, 

&2, 

C2 

and 

bu 

&2, 

63 

as, 

&3, 

C3 

Cl, 

C2, 

Cb 

in  each  of  which,  by  §  684,  the  expanded  form  is 

aiboCs  —  aihsCi  +  a^bsCi  —  a^hic^  +  as&iCg  —  a^biCi. 
Proof  of  the  theorem. 


484 


ADVANCED  COURSE  IN  ALGEBRA 


Consider  the  determinants 


^2,  IJ        ^h,  2? 


Otj 


f  2,  « 


and 


^1,1? 


^2,  IJ 
^2,  2J 


'n,2 


^«,  1?        ^n,  2j        '•'}       (^n,  n  ^1,  n?        ^2,  nj        •  •  •  j        <^n,  n 

Since  the  second  suffixes  of  the  first  determinant  are  the 
same  as  the  first  suffixes  of  the  second,  if  the  first  determinant 
be  expanded  by  the  rule  of  §  686,  and  the  second  by  the  rule 
of  §  690,  the  results  will  be  the  same. 

Therefore  the  determinants  are  equal. 

692.  A  determinant  is  changed  in  sign  if  any  two  consecutive 
rows,  or  any  tivo  consecutive  columns,  are  interchanged. 

The  truth  of  the  theorem  when  the  first  and  second  rows  of  a  determi- 
nant of  the  third  order  are  interchanged,  may  be  seen  by  expanding  the 
determinants 


and 


&2, 
&3, 


By  §  684,  the  expanded  forms  are,  respectively, 
«i&2C3  —  aihsC2  +  azbsCi  —  a^hiCs  +  a3&iC2 
and  a2&iC3  —  aa&sCi  +  ai&3C2  —  ai&2C3  +  asbiCi 

one  of  which  is  the  negative  of  the  other. 

Proof  of  the  theorem. 

Consider  the  determinants 


«3&2Cl, 

«3&iC2 ; 


''1,1? 


^r,  IJ        (^r 


^?,2? 


^M,1J 


and 


<^>-,  1>        <^r,  2) 


C('r,n 


*'?,!> 


^3,2? 


'^n.lj        ^n,2} 


Cln,n 


the  gth  and  rth  rows  of  the  first  being,  respectively,  the  rth 
and  gth  rows  of  the  second. 

Let  the  absolute  value  of  one  of  the  terms  of  the  first  deter- 
minant be  ^      ...  ^     n      ---a„„;  (1) 


'1,8 


<^5,«Ctr, 


where  s,  •••,  i,  w,  •••;  v  is  a  permutation  of  1,  2,  •••,  w, 


DETERMINANTS  485 

Since  the  element  in  the  qih  row  and  '^th  column  oi  the 
second  determinant  is  a,.^<,  and  the  element  in  the  rth  row 
and  uth  column  a^^  ^,  the  absolute  value  of  the  corresponding 
term  of  the  second  determinant  may  be  obtained  from  (1)  by- 
replacing  a^^t  and  a;,„  by  a^,j  and  a,,^,  respectively;  that  is, 

The  latter  expression  is  also  the  absolute  value  of  one  of  the 
terms  of  the  first  determinant,  since  it  has  one  and  only  one 
element  from  each  row,  and  one  and  only  one  from  each 
column ;  and  writing  it  so  that  the  first  suffixes  shall  occur  in 
the  order  1,  2,  •  •  •,  n,  we  have 

Cf'l,*"'Cl<l,uCi,,t"'Cln,v'  (2) 

Now  whatever  the  number  of  inversions  in  the  second  suf- 
fixes of  (1),  s,  '",  t,  u,  '",  V,  the  number  of  inversions  in  the 
second  suffixes  of  (2),  s,  •••,  u,  t,  •••,  v,  differs  from  it  by  unit}^; 
for  in  the  first  case  t  precedes  u,  and  in  the  second  u  pre- 
cedes t. 

Hence,  the  terms  (1)  and  (2)  of  the  first  determinant  are  of 
opposite  sign  (§  686). 

That  is,  any  two  terms  of  the  given  determinants  of  equal 
absolute  value  are  of  opposite  sign ;  and  hence  the  determi- 
nants themselves  are  of  equal  absolute  value  and  opposite  sign. 

It  follows,  from  §§  691  and  692,  that  if  two  consecutive 
columns  are  interchanged,  the  sign  of  the  determinant  is 
changed. 

693.  A  determinant  is  changed  in  sign  if  any  two  rows,  or  any 
two  columns,  are  interchanged. 

It  follows  from  the  proof  of  §  688  that  any  two  rows,  or 
any  two  columns,  of  a  determinant  may  be  interchanged  by  an 
odd  number  of  interchanges  of  consecutive  rows  or  columns. 

But  every  interchange  of  two  consecutive  rows  or  columns 
changes  the  sign  of  the  determinant  (§  692). 

Therefore,  the  sign  of  the  determinant  is  changed  if  any  two 
rows,  or  any  two  columns,  are  interchanged. 


486 


ADVANCED  COURSE  IN  ALGEBRA 


694.    Cyclical  Interchange  of  Rows  or  Columns. 

By  n—1  successive  interchanges  of  two  consecutive  rows, 
the  first  row  of  a  determinant  of  the  nth  order  may  be  made 
the  last. 

Thus,  by  §  692,  the  determinant 


TYli 
7112 


'nK 


is  equal  to  (— ly 


h, 

h} 

'") 

m< 

^n, 

K 

"') 

m, 

h, 

h, 

'") 

m^ 

The  above  is  called  a  cyclical  interchange  of  rows. 

In  like  manner,  hjn  —  1  successive  interchanges  of  two 
consecutive  columns,  the  first  column  of  a  determinant  of  the 
nth  order  msij  be  made  the  last. 

695.  If  two  rows,  or  two  columns,  of  a  determinant  are  iden- 
tical, the  value  of  the  determinant  is  zero. 

The  truth  of  the  theorem  when  the  first  two  rows  of  a  determinant  of 
the  third  order  are  identical,  may  be  seen  by  expanding  the  determinant 

«i,     hu     ci 
ai,     6i,     ci 

052,       &2,       C2 

By  §  684,  the  expanded  form  is 

ai&iC2  —  ai&2Ci  +  a\b2Ci  —  aibic^  +  a^hiCi  —  a^hiCi,  or  0. 

Proof  of  the  theorem. 

Let  D  be  the  value  of  a  determinant  having  two  rows,  or 
two  columns,  identical. 

If  these  rows,  or  columns,  are  interchanged,  the  value  of  the 
resulting  determinant  is  —  D  (§  693). 

But  since  the  rows,  or  columns,  which  are  interchanged  are 
identical,  the  two  determinants  are  of  equal  value. 

Hence,  D  =  —  D:  and  therefore  D  =  0. 


If  each  elemeiit  in  one  column,  or  in  one  row,  is  the  sum 
of  m  terms,  the  determinant  can  he  expressed  as  the  sum  of  m 
determinants. 


DETERMINANTS 


487 


The  truth  of  the  theorem  when  each  element  in  the  first  column  of  a 
determinant  of  the  third  order  is  the  sum  of  two  terms,  may  be  seen  by- 
expanding  the  determinant ;  consider,  for  example,  the  determinant 

ai  +  di,     &i,     ci 

CI2  +  d2,     62,     C2 

as  +  ds,     &3, 

By  §  684,  the  expanded  form  is 

(ai  +  di)  (&2C3  -  &3C2)  +  («2  +  d-z)  (&3C1  -  61C3) 
+  (as  +  ds)  (61C2  -  62G1) 
=  ai(&2C3  -  63C2)  +  a^CpsCi  -  biCs)  +  a3(biC2  —  62C1) 
+  di^bzCi  -  hzCi)  +  d2(hsci  -  &1C3)  +  dz(hiC2  -  62C1) 


ai, 

&i, 

Cl 

cZi, 

61, 

Cl 

«2» 

&2, 

C2 

+ 

^2, 

&2, 

C2 

as, 

&3, 

C3 

^S, 

&3, 

cs 

Proof  of  the  theorem. 
Consider  the  determinant 

^2,1?     •••> 


^2,rj 


^9   . 


(1) 


Let  each,  element  in  the  rth  column  be  the  sum  of  m  terms, 
ai,r=&i  +  Ci-h  •••  4-/1, 

a2,r=^2+C2H-    •••    +/2, 


as  follows : 


Let  tti.p  •••  ctj^r  •••  ^«,«  be  the  absolute  value  of  one  of  the  ele- 
ments of  (1) ;  then, 


Oi. 


•  a„ 


It  is  evident  from  this  that  the  determinant  (1)  can  be  ex- 
pressed as  the  sum  of  the  determinants 


al,l> 

a2,u 

"',      «l,n 
•",      (^2,n 

^n,l> 

-,   K, 

"}      Ctn,« 

+  •••  + 


«1,1) 

<^,  1) 


'  '>     Jni 


«2,n 


488 


ADVANCED  COURSE  IN  ALGEBRA 


697.  If  all  the  elements  in  one  column,  or  in  one  roio,  are 
multiplied  by  the  same  number,  the  determinant  is  multiplied  by 
this  number. 

The  truth  of  the  theorem,  when  all  the  elements  in  the  first  column 
of  a  determinant  of  the  third  order  are  multiplied  by  the  same  number, 
may  be  seen  by  expanding  the  determinant. 

Consider,  for  example,  the  determinant 

«2,      &2,      C2   .  (1) 

Multiplying  each  element  in  the  first  column  by  m,  we  have 


(2) 


mai, 

&i, 

Cl 

ma25 

62, 

C2 

mas, 

&3, 

cs 

The  expanded  form  of  (2)  is 

jnai(62C8  —  &3C2)  +  ma2{hsCi  —  &1C3)  +  wasC&iCa  —  boCi), 
which  is  m  times  the  expanded  form  of  (1). 

It  is  evident  from  this  that  (2)  equals  (1)  multiplied  by  m. 

Proof  of  the  theorem. 
Consider  the  determinant 


"'1,1? 

^2,1) 


}     ^l,r> 
'y     Ct'2,  r) 


f2,« 


^»i,l)      *•*>      ^n 


(1) 


Multiplying  each  element  in  the  rth  column  by  m,  we  have 

«1,1J      •••?     '^(^l,r,     ■-,     «l,n 
<^2,1)     •••>     ^^2,r?     •••?     ^2,w 


a„ 


ma„ 


(2) 


Let  Oj  p---  a^  ^••«  a„,  s  he  the  absolute  value  of  one  of  the  terms 
of  (1).    ■ 

Replacing  a^,^  by  ma^^rj  ^^^  absolute  value  of  the  corre- 
sponding term  of  (2)  is  mai.p---a^,^ •••«„,«• 

It  is  evident  from  this  that  the  determinant  (2)  equals  m 
times  the  determinant  (1). 


DETERMINANTS 


489 


698.  If  all  the  elements  in  any  column,  or  roio,  he  multiplied 
by  the  same  number,  and  either  added  to,  or  subtracted  from,  the 
corresponding  elements  in  another  column,  or  roio,  the  value  of 
the  determinant  is  not  changed. 

Let  the  elements  in  the  7'th  column  of  the  following 
determinant  be  multiplied  by  m,  and  added  to  the  correspond- 
ing elements  in  the  gth  column. 


(1) 


«1)     '",      «2J       •••>     <^rj      •",     «n 

K  •••?    K    "'^    ^r,   "-,    K 

, 

^IJ     •*•?      "'2?      *•*?      "'r?      *••?     '^n 

ain  the  determinant 

«i)    •••?   a^-\-ma„    •••,   a^,    •••,   a 

K    "-,   ^  +  ^i^r,    •••,    K     ",   &, 

„    ...,   k^  +  mK,    ..., 
which,  by  §§  696  and  697,  is  equal  to 


(2) 


6i, 


7c„ 


iCj. 


+  m 


ki, 


K 


But  the  coefficient  of  m  is  zero  (§  695). 
Whence,  the  determinant  (2)  is  equal  to  (1). 

699.   Minors. 

If  the  elements  in  any  m  rows  and  any  m  columns  of  a 
determinant  of  the  nth.  order  be  erased,  the  remaining  ele- 
ments form  a  determinant  of  the  (n  —  m)th  order. 

This  determinant  is  called  an  mth  Minor  of  the  given  deter- 
minant. 


Thus, 


ai, 

di, 

ei 

Ct's, 

^3, 

63 

«6, 

^5, 

65 

is  a  second  minor  of 


^1}  ^1}  ^1}  ^i>  ^1 

dgj  t>2)  ^2>  ^*2j  ^2 

%>  ^3>  ^3)  ^3>  ^3 

<*4j  ^4>  <^4j  •^4?  ^4 

^5)  ^5;  ^5?  ^5>  ^5 


490 


ADVANCED  COURSE  IN  ALGEBRA 


It  is  obtained  by  erasing  the  second  and  fourth  rows,  and 
the  second  and  third  columns,  of  the  latter  determinant. 


700.    To  find  the  coefficient  of  a^  ^  in  the  determinant 


^1,  IJ      ^1,  2) 
^2,  1?      ^2,  2J 


^2,  n 


\  1)      ^n,  2j 


(1) 


By  §  686,  the  absolute  values  of  the  terms  which  involve 
ai^i  are  obtained  by  forming  all  possible  products  of  the  ele- 
ments taken  n  at  a  time,  subject  to  the  restrictions  that  the 
first  element  shall  be  ai^  i,  and  that  each  product  shall  contain 
one  and  only  one  element  from  each  row  except  the  first,  and 
one  and  only  one  from  each  column  except  the  first. 

It  is  evident  from  this  that  the  coefficient  of  aj^i  in  (1)  may 
be  obtained  by  forming  all  possible  products  of  the  following 
elements  taken  ti  —  1  at  a  time, 


^2, 2?      ^2, 
^3, 2j     ^3, 


^2,  n 


subject  to  the  restriction  that  each  product  shall  contain  one 
and  only  one  element  from  each  row,  and  one  and  only  one  from 
each  column,  writing  the  first  suffixes  in  the  order  2,  3,  •••,  n, 
and  making  each  product  +  or  —  according  as  the  number  of 
inversions  in  the  second  suffixes  is  even  or  odd. 
Then  by  §  686,  the  coefficient  of  a^^  j  is 

2,2)     ^2,3)      ••*?     <^2,  i 


1,  2)      ^n,  3>      *  *  *)     "n 

that  is,  the  minor  obtained  by  erasing  the  first  row  and  the 
first  column  of  the  given  determinant. 

701.   By  aid  of  §  700,  a  determinant  of  any  order  may  be 
expressed  as  a  determinant  of  any  higher  order. 


DETERMINANTS 


491 


Thus, 


ttl, 

K 

Ci 

«2, 

h, 

C2 

= 

«3, 

K 

C3 

1, 

0,   0,   0 

0, 

tti,  bi,  Ci 

0, 

a2,  h,  C2 

0, 

as,  bsy  C3 

1,  0,  0,  0,  0 

0,  1,  0,  0,  0 

0,  0,  ai,  61,  Ci 

0,  0,  a^,  62,  C2 

0,  0,  as,  bs,  Cg 


etc. 


702.   Coefficient  of  any  element  of  a  Determinant. 
To  find  the  coefficient  of  63,  in  the  determinant 


Oj, 

h, 

Ci 

ttg, 

h, 

C2 

ttg, 

K 

C3 

(1) 


By  two  interchanges  of  consecutive  rows,  the  last  row  may 
be  made  the  first ;  thus,  by  §  6y2,  the  determinant  equals 


(-ly 


^3, 

h,  C3 

a„ 

&1,    Ci 

a^. 

h,  C2 

(2) 


By  interchanging  the  first  two  columns,  the  determinant 
(2)  equals 


(-1)= 


By  §  700,  the  coefficient  of  b^,  in  (3),  is 


&3, 

«3, 

C3 

K 

Oi, 

Cl 

K 

«2, 

C2 

(3) 


(-ly 


ttj,    Ci 


That  is,  the  coefficient  of  the  element  in  the  third  row  and 
second  column  equals  (—1)^+^  multiplied  by  that  minor  of  (1) 
which  is  obtained  by  erasing  the  third  row  and  second  column. 

We  will  now  consider  the  general  case ;  to  find  the  coefficient 
of  a^fc,  f  in  the  determinant 


(4) 


ttl.lJ 

'"} 

«l,r^ 

'"} 

(h,n 

«*.l, 

'"} 

«*,rj 

'*•> 

Ctk.n 

«»,  1} 

'  "i 

Ctn,r, 

'") 

«n,«/ 

492 


ADVANCED  COURSE   IN  ALGEBRA 


By  k  —  1  interchanges  of  consecutive  rows,  and  r  —  1  inter- 
changes of  consecutive  columns,  the  element  a^^,.  may  be 
brought  to  the  upper  left-hand  corner. 

Thus,  by  §  692,  the  determinant  equals 


(- !)'-'(- ir' 


^k.n 


a. 


a„ 


Then,  by  §  700,  the  coefficient  of  a^^  ^  is 


^1,1? 


1,  » 


''n,  17 


But 


Hence,  the  coefficient  of  the  element  in  the  Z:th  row  and  rth 
column  equals  (—  l)*"^*",  multiplied*  by  that  minor  of  (4)  which 
is  obtained  by  erasing  the  A:th  row  and  7'th  column. 

703.  By  aid  of  §  702,  a  determinant  of  any  order  may  be 
expressed  in  terms  of  determinants  of  any  lower  order. 

Thus,  since  every  term  of  a  determinant  contains  one  and 
only  one  element  from  the  first  row,  we  have, 


^4, 


&1, 
&2J 


=  ai 


+  Ci 


b„ 

Ci, 

(k 

a,, 

Cl, 

d. 

h, 

Cs, 

d. 

-h 

«3, 

Cs, 

d. 

K 

Ci, 

d. 

«« 

c« 

d. 

<h, 

\, 

d. 

«» 

h, 

C2 

as, 

K 

d. 

-d. 

«» 

K 

Cs 

«4, 

K 

d. 

a,, 

K 

C4 

and  each  of  the  latter  determinants  may  in  turn  be  expressed 
in  terms  of  determinants  of  the  second  order. 

704.   Evaluation  of  Determinants. 

The  method  of  §  703  may  be  used  to  express  a  determinant 
of  any  order  higher  than  the  third  in  terms  of  determinants 
of  the  third  order,  which  may  be  evaluated  by  the  rule  of  §  684. 


DETERMINANTS 


493 


1.  Evaluate 


The  theorem  of  §  698  may  often  be  employed  to  shorten  the 
process,  as  shown  in  Ex.  1. 

5,       7,       8,--  6 

11,     fe,     13,     11 

14,     2'4:,     20,     23 

7,     13,     12,      2 

Subtracting  the  first  row  from  the  last,  twice  the  first  row 
from  the  second,  and  three  times  the  first  row  from  the  third 
(§  698),  the  determinant  becomes 


by  §  697. 


Subtracting  five  times  the  second  row  from  the  first,  adding 
the  second  row  to  the  third,  and  subtracting  the  second  row 
from  the  last,  we  have 


5, 

7, 

8, 

6 

5, 

7, 

8, 

6 

1, 

1, 

2, 
3, 

-3, 
-4, 

-1 
5 

=  2 

1, 
-1, 

2, 
3, 

-3, 
-4, 

-1 

5 

2, 

6, 

4, 

-4 

1, 

3, 

2, 

-2 

0, 

-3, 

23, 

11 

1, 

.2, 

-3, 

-1 

0, 

5, 

-7, 

4 

0, 

1, 

5, 

-1 

-3, 

23, 

11 

=  -2 

5, 

-7, 

4 

h 

5, 

-1 

,  by  §  702. 


The  object  of  the  above  process  is  to  put  the  given  deter- 
minant in  such  a  form  that  all  but  one  of  the  elements  in  one 
column  shall  be  zero ;  the  determinant  can  then  be  expressed 
as  a  determinant  of  the  third  order  by  §  702. 

The  last  determinant  may  be  evaluated  by  §  684 ;  but  it  is 
better  to  subtract  five  times  the  first  column  from  the  second, 
and  then  add  the  first  column  to  the  last ;  thus, 
-  3,      38,  8 


32,9 
0,0 


38,8 
-32,9 


=  -  2(342  +  256)  =  - 1196. 


2.   Prove  that  a  +  &  +  c  is  a  factor  of  the  determinant 


a, 

b, 

c 

c, 

a, 

b 

^ 

C; 

a 

494 


ADVANCED  COURSE  IN  ALGEBRA 


3.  Evaluate 


Adding  the  second  and  third  columns  in  succession  to  the 
first,  the  given  determinant  equals 

a-\-b-\-c,     b,     c 
c-\-a-{-b,    a,     b 
b-\-c-^a,     c,    a 
Since  every  term  contains  an  element  from  the  first  column, 
a-i-b-\-c  is  a  factor  of  the  given  determinant. 

«v*  Jb   m  Jb 

y,   y%   f 

If  X  be  put  equal  to  y,  the  determinant  has  two  rows  identical, 
and  equals  zero  (§  695). 

Then,  x  —  y  must  be  a  factor  (§  180) ;  and  in  like  manner, 
y  —  z  and  z  —  x  are  factors. 

Let  the  given  determinant  =  X(x  —  y)(y  —  z)(z  —  x). 

To  determine  X,  we  observe  that  x,  y,  and  z  are  factors  of 
the  determinant;  then,  X  must  equal  xyz,  as  is  evident  by 
noticing  that  the  first  term  in  the  expanded  form  is  -{-xyh^^ 
and  the  value  of  the  determinant  is  xyz(x  —  y){y  —  z)(z  —  x). 


■ 

EXERCISE 

115 

Evaluate  the  following  determinants  : 

7,      8,      9 

1,    a,    «» 

1. 

28,     85,    40 
21,     26,     30 

9,     13,     17 

•     4. 

1,     &,     68 
1,     c,     c8 

1,      1,      1 

7. 

2. 

11,     15,     19 
17,    21,     25 

.     5. 

a;2,     ?/2^     2;2 
x^,    y^    03 

8. 

10. 


25,    23,     : 
14,     11, 
21,     17,     : 

a,  6, 

&,  -a, 

c,  0, 

0,  c, 


1,  1, 

6,  11, 

8,  7, 

1,  6, 


1, 

25, 
1, 
6, 


c, 

0 

iC,   x,  X,   X 

0, 

a. 

c 
h 

11. 

X,  y,  X,  X 
X,  X,  y,  X 

12. 

&, 

—a 

X,  X,  X,  y 

5 

5,      2 

'? 

14,      6 

'l 

1,      4 

), 

21,       9 

*? 

11,     10 

^ 

12,       9 

» 

10,     11 

M 

9,     12 

s 

5,       9 

, 

8,     11 

'» 

4,      2 

> 

5,     10 

a, 

a,  6,  a 

&, 

6,  6,  a 

&, 

a,  a,  a 

&, 

a,  &,  & 

DETERMINANTS 


495 


13. 


.:  14. 


i> 


0,  1,      1,      1 

1,  0,      a2,     62 
1,     a\    0,     c2 

1,       b\       C2,       0 

• 

15. 

7,     10,     13,      3 
14,     19,     27,       6 
24,    33,     41,     10 
31,    47,     64,     15 

16. 

5,   -3,   -2, 

4,       1,   -6, 

-  1,       4,       3, 

0,       6,   -4, 

0 

2 

-5 

2 

1,      1,      1,      1 
a,     6,      c,      d 

a2,     62^      c2,      C?2 
a3,    63,     c3,     # 

17.   Prove  that  a  +  6  +  c  is  a  factor  of  the  determinant 


(a  +  6)2, 


62, 


(6  +  c)2, 


18.   Prove  that  a^b  -\-c 


b^       (c  + 

ay 

-d 

is  a  factor  of  the  determinant 

a, 

6,       c,  -d 

h 

a,  —  d,      c 

c, 

—  d,       a,      6 

d, 

c,       6,      a 

705.  Let  A,,  B„ 


/r^,  denote  the  coefficients  of  the  ele- 


ments ar,  b^, 


-,  k^,  respectively,  in  the  deterjninant 
^ij     ^1)     '"}     ki 

«£,         O2,         "•,         K"2 


a„,    b^,     •••,     lc„ 


(1) 


Then,  since  every  term  of  the  determinant  contains  one  and 
only  one  element  from  the  first  column,  the  value  of  the  deter- 
minant is  ^^^^  ^  ^^^^  H-  . . .  +  A^a,. 

In  like  manner,  the  value  of  the  determinant  also  equals 

706.  With  the  notation  of  §  705,  if  m^  m^,"-,  m„  are  the 
elements  in  any  column  of  the  determinant  (1),  of  §  705,  except 
the  first,  ^^^^  _^  ^^^^  +  . .  •  +  Awn 

is  the  value  of  a  determinant,  which  differs  from  (1)  only  in 
having  mi,  mg,  •••,  m„  instead  of  a^,  a^,  ••♦,  a„  as  the  elements  in 
the  first  column. 


496 


ADVANCED  COURSE  IN  ALGEBRA 


Then,  Ainii  +  ^2^2  H h  ^n^n  =  0 ; 

for  it  is  the  value  of  a  determinant  which  has  two  columns 
identical. 

In  like  manner,  if  mj,  mg,  •••,  m„  are  the  elements  in  any 
column  of  the  determinant  (1),  except  the  second, 

BiTrii  +  ^2^2  H h  -S„m„  =  0 ; 

and  so  on. 

SOLUTION  OF  EQUATIONS 

707.   Let  it  be  required  to  solve  the  following  system  of  n 
linear,  simultaneous  equations,  involving  n  unknown  numbers  :^ 
aiXi  +  ...  +  g^aj,  H h \x^  =Pi-  (1) 


Cin^l  +  •••  +  gn«r  H h  K^n  =  Pn-         ■  (3) 

Let  Qi,  Qs?  •••?  Qn  denote  the  coefficients  of  the  elements  qi, 
Qi)  '"}  Qn}  respectively,  in  the  determinant 

^i>  ••*>  9ij  '"}  ^1 

jr^__     <^2>    •••?  Q2)    '"}  1^2 


Multiplying  equations  (1),   (2),  ••.,  (3)  by  Q^   Qa,   —,  Q„, 

respectively,  and  adding,  we  have 

Xi(QiC('i  +  Q2«2  +  •••  +  Q„a„)+  ... 
+  ^XQiQi  +  ^292  4-  •••  +  Qn9»)+  - 

4-aJ„(QA  4-   ©2^2  +  •••  +  QnK)=  QlPl  +  Q2P2  +  -  +  QnPn- 

By  §  706,  the  coefficient  of  each  unknown  number,  except  x^, 
is  zero. 

By  §  705,  the  coefficient  of  x^is  D;  also,  the  second  member 
is  the  value  of  a  determinant  which  differs  from  D  only  in 
having  p^,  p2,  •••,  p„  instead  of  q^,  q^,  -",  g„  as  the  elements  in 
the  rth  column. 

Denoting  the  latter  by  Z>^,  we  have 

A 


x^D  =  D^,  and  x^  = 


D 


DETERMINANTS 


497 


The  determinant  D  is  called  the  Discriminant  of  the  given 
system  of  equations. 

708.    Ex.   Find  the  value  of  y  from  the  equations 

2a;  +  6?/-92=-23. 
.4£c  —  2?/  —  52;=        9; 

The  denominator  of  the  value  of  y  is  the  determinant 

3,  -6,       7 
2,       6,    -9 

4,  -2,    -5 

The  numerator  is  obtained  by  putting  for  the  second  column 
the  second  members  of  the  given  equations. 


Therefore, 


3, 

28, 

7 

2, 

-23, 

-9 

4, 

9, 

-5 

3, 

-   5, 

7 

2, 

6, 

-9 

4, 

-   2, 

-5 

630 


210 


=  -3. 


709.   Consider   the    following    system   of   n   homogeneous 
linear  equations,  involving  n  unknown  numbe^rs: 

aja^i  +  h^X2  +  CiO^g  H h  \x^_^  +  \x^  =  0. 

a.jXi  +  b^2-{-c^3-\ h  M«-i  +  ^2aJ«  =  0.  (1) 


««a;i  +  b^X2 -\-CnXs-\ h  KX^_-^  +  A^n^^n  =  0. 

One  solution  of  the  system  is  a;i  =  0,  iC2  =  0,  •••,  a/'„  =  0- 
We  will '  now  find  what  relations  must  hold  between  the 
coefficients  in  order  that  there  may  be  other  solutions. 
We  may  write  the  given  system 

b^x^  4-  Ci»3  H h  Mn  =  -  (h^i- 

b^2  +  c^3-^ h  Mn  =  -  Clffh'  (2) 

[  bnX2  +  Cn^g  H h  Kx^  =  —  a^^x^. 

We  will  now  use  the  last  71  —  1  of  these  equations  to  express 
the  values  of  X2,  •••,  x^  in  terms  of  Xy. 


498 


ADVANCED  COURSE  IN  ALGEBRA 


By  §  707, 

0-2^1,       C2, 


—  a^x^,    c. 


•)        K,n 


^25        ^2?         •••>        "^2 


^n»        ^nJ       '"j        "'n 


=  -iCl 


a» 

c» 

-,     *2 

«M 

c« 

-,    K 

6„ 

Cj, 

...,  h. 

6« 

c« 

■■;    K 

with  results  of  similar  form  for  iCg,  •••,  x^. 

Substituting  these  values  in  the  first  equation  of  system  (1), 
and  clearing  of  fractions,  we  have 


O2,    C2,    •••,    A;2 

^n5     ^n?     *••?     "^n 


-&1 


A^l 


^2)      ^2,       •••,      fcj 

^n)      ^n)       '*•?      "'m 
^2^        ***>       "'2?       ^2 


—  C, 


62J      0^2,      ••*,      /C2 


^n)      ^n)      •••>     \ 


'n>  7      '"nJ      "'n 


By  §  693,  the  coefficient  of  x^  equals 


0. 


(3) 


Oj 


K 

•••,   Zcg 

• 

.     . 

-*i 

K 

-,  K 

«2,    C2, 

•••,  k2 

•        • 

'    ' 

+  Ci 

««,  c,, 

-,  A:. 

^2j    ^2j    ^2>     •**;    "'2 
^nj    ^»)    ^n>    ••*>    "'n 


That  is  (§  702), 


^IJ       ^1,       •••,       /Cj 
<^2J        ^2>        •*•?       "^2 


^n>       ^n?        '")       ""n 


(4) 


If  the  determinant  (4)  equals  0,  equation  (3)  is  satisfied  when 
a^  has  any  real  value. 

Then,  if  the  discriminant  of  the  given  system  vanishes,  the 
system  has  an  indefinitely  great  number  of  solutions. 

If  any  real  number,  m,  be  taken  as  a  value  of  xi,  the  corresponding 
value  of  X2  must  be 


a2, 

C2,       • 

•,     k2 

an, 

Cn,       • 

;     Tcn 

H 

C2,       •• 

.,    k2 

bn, 

Cn,       • 

.,      kn 

DETERMINANTS 


499 


Similar  considerations  hold  with  respect  to  xs,  •••,  a;„. 

710.   Consider  the  following  system  of  n  linear  equations 
involving  n  —  1  unknown  numbers  : 
■b  r  aiXi ■i-biX2-\ h  QiXn-i  +  A^i  =  0. 

^■k  I   a2a^l  +  &2a'2  H h  g2^n-l  +  ^2  =  0. 

[anXi+bnX2-\ h  gn^^n-l  +  ^»  =  0. 

Multiplying  the  terms  of  each  equation  by  u,  we  have 

a^uxi  +  biUX2  H h  q\ux,^  -^  +  TciU  =  0. 

aswo^i  +  62^^*2  H h  ^2^''^n-i  +  A;2?t 


0. 


(1) 


a^WiCj  -h  &„Wa72  H h  gnU^n-l  +  ^n^  =  0.  " 

This  may  be  considered  as  a  system  of  n  homogeneous  linear 
equations,  involving  the  n  unknown  numbers,  ux^,  ux2,  •••, 
ux^^i,  u. 

By  §  709,  the  condition  which  must  hold  in  order  that  the 
system  (1)  may  have  possible  solutions  is 

%j    bi,     '••}     5'ij    "'I 


Q2}      1^2 


Kj      ^n) 


fc^ 


=  0. 


This,  then,  is  the  condition  which  must  hold  in  order  that  the 
equations  of  the  given  system  may  be  consistent. 

711.   Multiplication  of  Determinants. 

Consider  the  determinant 

ttiCj  +  a2C?i,   biCi  4-  62^1 

aiC2  -\-  a^d^,    61C2  +  62^2 
By  §  696,  this  can  be  expressed  in  the  form 


4- 


Orfil,       61C1 

^1^2?      ^1^2 

That  is,  by  §  697, 

Ci,     Ci 

C2,   C2 


a.c. 


b.A 

(XjC2,     02^2 


a^d^,    biCi 

^2^2)      ^1^2 


4- 


aodi,   62^1 
a2d2,   M2 


aj6i 


H-«A 


Cg?    ^2 


+  «2^1 


(X2,     C2 


+  a2&2 


0(2)     ^2 


600 


ADVANCED  COURSE  IN  ALGEBRA 


The  first  and  last  determinants  are  zero,  since  they  have  two 
columns  identical. 

Then,  by  §  692,  the  given  determinant  equals 

a-p2 


C2,     d2 

-aafti 

Ci,   d^ 

Cg,     ^2 

=  («! 

^2-«2&] 

) 

C2,     ^2' 

= 

a2,   &2 

X 

C2,     ^2 

That  is,  the  product  of  two  determinants  of  the  second  order 
can  be  expressed  as  a  determinant  of  the  second  order. 
Again,  consider  the  determinant 

ajdi  +  asCi  +  tts/i,   Ml  +  ^261  +  &3/1J   Mi  +  c^ei  +  Cs/i 
aid2  4-  ^262  +  ^sA   hdi  +  62^2  +  &3/2J   CiC^2  +  ^262  +  Cg/g 

«l<^3  +  ^263  +  tts/s;      ?^1^3  +  ^2^3  +  ^3/3?      ^1^3  +  0263  +  03/3 

By  §  696,  this  equals  the  sum  of  twenty-seven  determinants, 
of  which  the  following  are  types : 


V11 

C3/1 

^2^2? 

C3/2 

) 

^2^3, 

C3/3 

Ci\d2, 

That  is,  by  §  697, 
d\)     ^ij    f\ 

0102^3     ^2)        ^2>      .^ 
d^y        ^3?       .^ 


Ct](X2y 

ctids, 


CfrfilC2 


hdi, 
bid2, 
hds, 


di, 


0261 

02^2 

CoCo 


dfi. 

61 

d2, 

^2 

ds, 

63 

1     1> 

Oidg, 
aidg, 


ai^iCj 


bid2, 
bids, 


C-^Oj2 

Cids 


di, 

di, 

di 

d2, 

d2, 

do 

ds, 

ds, 

ds 

The  eighteen  determinants  of  the  second  type,  and  the  three 
of  the  third  type,  are  all  zero  by  §  695. 
Hence,  the  given  determinant  equals 


ciihcs 


+  azbsCi 


di,    61, 

/i 

di,   fi,    e. 

ei,    d„   /i 

^2,      62, 

/2 

+  «1?>3C2 

d„    U    e.2 

+  a2&iC3 

62,      ^2,      /s 

^3,    eg, 

/s 

ds,    fs,    63 

637      ^3J     J^ 

ei,    fi, 

d. 

A,    ^ij    ei 

fi,    ^1,    di 

^2,      ./2) 

do 

+  a,b^C2 

.4      ^2J      62 

+  (^sbiOi 

f2,      ^2,      d2 

es,    f„ 

ds 

/sj    d^,    63 

fi,      63,       dg 

DETERMINANTS 


601 


By  §  692,  the  abpve  equals 


or, 


&1, 

Ci 

&2, 

C2 

X 

?>3, 

C3 

+  ag&iCg  — 

ag^^Ci) 

(^,    61,   /i 

^2J      62,     /g 

• 

C^3J       63?     /3 

^i, 

ei,   /i 

d^, 

^2,      /2 

ds, 

63,      /3 

That  is,  the  product  of  two  determinants  of  the  third  order 
can  be  expressed  as  a  determinant  of  the  third  order. 


EXERCISE  116 


Solve  the  following  by  determinants  : 

x  +  2y  -\-Ss=   4. 
1.  \  3z+5y+    z  =  18.  3. 

4a;  +    y-\-2z  =  12. 


4. 


8x-3y-  7;^  = 

85. 

x  +  6y-4z  =  - 

-12. 

2x  —  6y+    z  = 

33. 

6.   Express 

4,    -3 

-7,        8 

X 

5,    -6 

10,        7 

1,        2,-6 

4,    - 

6.   Express 

0,-1,        1 

X 

-1,    - 

2,    -3,    -5 

-7, 

5,        0,        1 

>^4'" 

7.   Express 

-5,        2,-1 

-1,    -8, 

41 

*> 

X  +     y+    z  =  l. 

ax  +  by  +  cz  =  d. 
a^x  +  b^y  4-  C^z  =  <^. 

«+    y+    2+    M=       1. 

2x  +  Sy-4-z  +  5w=-  31. 

3x-4y  +  52;  +  6w=-22. 

.4x  +  5y-60-    u=:-13. 

as  a  determinant. 


1,    -3 
4,    -5 

0,        2 


as  a  determinant. 


8.   Express  the  square  of 


a,  5,  c 
6,  c,  a 
c,  a,  6 


3  I  as  a  determinant. 
5  I      pare  §  701.) 


a  determinant. 


(Com- 


502       ADVANCED  COURSE  IN  ALGEBRA 


XXXVII.    THEORY  OF  EQUATIONS 

712.  Every  equation  of  the  nth  degree,  with  one  unknown 
number,  can  be  written  in  the  form 

where  the  coefficients  may  be  positive  or  negative,  integral  or 
fractional,  rational  or  irrational,  real,  pure  imaginary,  or  com- 
plex, or  zero. 

If  no  coefficient  equals  zero,  the  equation  is  said  to  be  CoTn- 
plete;  otherwise,  it  is  said  to  be  Incomplete. 

We  shall  call  (1)  the  General  Form  of  the  equation  of  the 
nth  degree. 

713.  It  will  be  proved  in  the  Appendix  that  every  equation 
of  the  above  form  has  at  least  one  root,  real,  pure  imaginary,  or 
complex. 

714.  (Converse  of  §  183.)     If  a  is  a  root  of  the  equation 

X^+P^X^-^  +  ...  +Pn-iX  +P.  =  0,  (1) 

the  first  member  is  divisible  by  x  —  a. 

If  a  is  a  root  of  (1),  a"  +Pia**"^  H hi>„  =  0. 

But  by  §  139,  a"+pia**~^  H hPn  is  the  remainder  obtained 

by  dividing  the  first  member  of  (1)  by  a;  —  a. 

Then,  the  first  member  of  (1)  is  divisible  by  a;  —  a. 

715.  Number  of  Roots. 

An  equation  of  the  nth  degree  has  n  roots,  and  not  more  than  n. 
Let  the  equation  be 

X^  4-  PlX""-"-  +P2X--'  +  •  •  •  -\-Pn-lX  +Pn  =  0.  (1) 

By  §  713,  this  equation  has  at  least  one  root. 
Let  a  be  this  root ;  then  by  §  714,  the  first  member  is  divisible 
by  ic  —  a,  and  the  equation  may  be  put  in  the  form 

{x  -  a)  (x--'  +  q^--""  +  . . .  +  qn-ix  +  g„)  =  0. 


THEORY   OF   EQUATIONS  603 

By  §  182,  the  latter  equation  may  be  solved  by  placing 
a;  —  a  =  0, 
and  x^-^  +  q^^-''  +  •  •  •  +  q^-xX  +  g,  =  0.  (2) 

Equation  (2)  must  also  have  at  least  one  root. 
Let  h  be  this  root ;  then  (2)  may  be  written 

{x  -  h)  (x--'  +  r,x--'  +  -  +  r^-ix  +  r,)  =  0, 
and  the  equation  may  be  solved  by  placing 

x-b  =  0, 
and  x**~^  +  rgic"-^  +  •••  +  r^^iX  +  r^  =  0. 

Continuing  the  process  until  7i  —  1  binomial  factors  have 
been  divided  out,  we  shall  arrive  finally  at  an  equation  of  the 
first  degree,  ^j  -  A;  =  0 ;  whence,  x  =  k. 

Therefore,  the  given  equation  has  the  7i  roots  a,  b,  •••,  k. 
The  roots  are  not  necessarily  unequal;  compare  note,  §  183. 

716.   Depression  of  Equations. 

It  follows  from  §  715  that,  if  m  roots  of  an  equation  of  the 
nth  degree  are  known,  the  equation  may  be  depressed  to 
another  equation  of  the  (n  —  m)th  degree,  which  shall  contain 
the  other  n  —  m  roots. 

Thus,  if  all  but  two  of  the  roots  of  an  equation  are  known, 
these  two  may  be  obtained  from  the  depressed  equation  by  the 
rules  for  quadratics. 

-'  Ex.   Two  roots  of  the  equation  9  x^  -  37  ic'  -  8  a;  +  20  =  0  are 
2  and  —  f ;  what  are  the  others  ? 

By  §  714,  the  first  member  of  the  given  equation  is  divisible 
by  {x  -  2)  (3a;  +  5),  or  3aj2  -  a;  -  10. 

Dividing  9  «*  -  37  a^  -  8  a;  +  20  by  3  ar^  -  a;  -  10,  the  quotient 
is3a^4-a:-2. 

Then  the  depressed  equation  is 

3a;2  +  a;-2  =  0. 

2 
Solving  by  the  rules  for  quadratics,  x  =  -  or  —  1. 

o 


504  ADVANCED  COURSE  IN  ALGEBRA 


EXERCISE  1!7 

1.  One  root  of  ic^  -  a:2  -  32  x  +  60  =  0  is  5  ;  find  the  others. 

2.  One  root  of  3  x^  -  6  x^  -  34  a: +  24  =  0  is  -  3  ;  find  the  others. 

3.  One  root  of  24  a;^  -  60  a;2  +  9  x  +  20  =  0  is  -  - ;  find  the  others. 

4.  One  root  of  27  x^  +  54  a;^  -  141  x  +  28  =  0  is  ^ ;  find  the  others. 

5.  Two  roots  of  9  x*  +  6  x*  -  69  x^  -  20  x  +  100  =  0  are  -.2  and  ^; 
find  the  others. 

6.  Two  roots  of  20  x*  -  177  x^  +  256  x^  +  492  x  +  144  =  0  are  4  and 
3 

;  find  the  others. 

4' 

7.  Two  roots  of  x*  +  ax^  -  22  a^x^  _  16  aH  +  96  a*  =  0  are  -  3  a  and 
4  a  ;  find  the  others. 

8.  One  root  of 

a;3  _^  (•^  +  6)  a;2  _  (^2  _  4  ^  _  11)  ^  _  ^3  _  2  n2  +  5,71  +  6  =0  is  -  w  -  3  ; 
find  the  others. 

717.   Formation  of  Equations. 

It  follows  from  §  715  that  if  the  roots  of 

are  a,  6,  •••,  h,  the  equation  may  be  written  in  the  form 
{x  —  a){x  —  'b)  "' {x  —  h)  =  0. 

Hence,  to  form  an  equation  which  shall  have  any  required 
roots, 

Subtract  each  root  from  x,  and  place  the  product  of  the  resulting^ 
expressions  equal  to  zero. 

Ex.     Form   an    equation   which    shall    have   the    roots    1, 

-,  and  — -• 

2'  3 


By  the  rule,        (x -l)(x- ^  fx-\-^=: 0. 


Multiplying  the  terms  of  the  second  factor  by  2,  and  of  the 
third  by  3,  we  have 

(x-T)(2x-l)(Sxi-5)=0. 
Expanding,  6  x^  +  x'-12  x  +  5  =  0. 


THEORY  OF  EQUATIONS  505 


EXERCISE  118 

Form  equations  having 

the  roots : 

1     ^     -7. 
^'   5'        6 

8.    4,  4,  -1    -1 

2.  1,   -4,  6. 

3.  -2±V'6. 

9.        m,   '^i/^ 
4 

4.    2,   -  3,  5,  0. 

10.    3±V2,   -3±V2,  0. 

5.  2  a  -  5,  3  a  +  4. 

6.  -1,   -2,  7,  -8. 

11.   a,  -b,  ^,    -I- 

a        b 

'•  ^' «'  1'  1- 

.„    4±2V3     -2±V3 
3       '           3        ■ 

18   2V-: 

2±\/-5 

_2\/-2±\/-5 

2 

2 

14.    n  +  1, 

-n-2, 

-71  +  3,  ?i  -  4. 

718.   Composition  of  Coefficients. 

By  §  717,  the  equation  whose  roots  are  a  and  b  is 

(x  —  a)(x  —  b)  =  0,  or  x^—(a-\-  h)x -\-ab  =  0. 
The  equation  whose  roots  are  a,  b,  and  c  is 
(x  —  a){x  —  b)(x  —  c)=Oy 
or  ic^— (a  +  6+c)x^+(a6  +  ac4-6c)a;  — a&c  =  0. 

The  equation  whose  roots  are  a,  b,  c,  and  d  is 
(x  —  a)(x  —  6)(ic  —  c)(£c  —  d)  =  0,  or 
iB*  —  (a  +  6  +  c  +  d)ar^  +  (ab  -{- ac -^  ad -\- be  -\-bd  +  cd)x^ 
—  (abc  +  aftd  +  acd  +  6cd)a;  +  abed  =  0. 
In  the  above  expanded  forms,  we  observe  the  following  laws : 
T7ie  coefficient  of  the  second  term  is  equal  to  minus  the  sum 
of  all  the  roots. 

TJie  coefficient  of  the  third  term  is  equal  to  the  sum  of  the 
products  of  the  roots,  taken  two  at  a  time. 

The  coefficient  of  the  fourth  term  is  equal  to  minus  the  sum 
of  the  products  of  the  roots,  taken  three  at  a  time;  etc. 
I      The  last  term  is  equal  to  plus  or  minus  the  product  of  all  the 
i  roots,  according  as  the  number  of  roots  is  even  or  odd. 


^^  506  ADVANCED  COURSE   IN   ALGEBRA 

-^        We  will  now  prove   by  Mathematical  Induction  that  the 
above  laws  hold  for  any  number  of  roots. 

Assume  them  to  hold  for  n  roots ;  a,  h,  c,  d,  •••,  k,  I,  m. 
That  is,  (x  —  d)(x  —  b)(x  —  c)  •••  {x  —  m) 

=  x'''-\-p^x''-^-{-p^''-^-\-p^x''-^-\-  •..  4-Pn; 
where     pi=—  (a  +  64-c+  •••  4-Zc  +  ?4-m); 
p^z=  ah  -{-  ac  -\-  he  -\-  •••  -\-lm\ 
Ps—  —  (ahc  -\-  ahd  +  acd  +  •  •  •  +  Mm)  ; 


Pn=  ±  <^hcd  •••  kbn,  according  as  n  is  even  or  odd. 
If  we  introduce  an  additional  root,  r,  we  equate  to  zero  the 
product  of 

x^ -\-piX''~'^  +^2^"*"^  +P3X^~^  +  •••  -\-Pn  by  x  —  r. 
This  gives  the  equation 

Here,  the  coefficient  of  the  second  term  is 

-(a  +  &  +  c+  •••  +A;+Z  +  m  +  r). 
Of  the  third  term 

ah -{- ac ■\- he -\-  •••  -{-Im  +  ria  +  h  -\-c-\-  •••  +m). 
Of  the  fourth  term 

—  {ahc-\-  •••  -\-ldm) —  7'{ah-\-aG-\-  •••  +  ^m). 


The  last  term  is  =F  ahcd  •  •  •  mr,  according  as  n  is  even  or  odd ; 
or  ±  ahcd  -•-  r,  according  as  n  + 1  is  even  or  odd. 

These  results  are  in  accordance  with  the  above  laws. 

Hence,  if  the  laws  hold  for  n  roots,  they  hold  for  ?i-|-l  roots. 

.But  we  know  that  they  hold  for  four  roots,  and  hence  they 
hold  for  five" roots;  and  since  they  hold  for  five  roots,  they 
hold  for  six  roots  ;  and  so  on. 

Hence,  the  laws  hold  for  any  number  of  roots. 

'  719.     It  follows  from  §  718  that,  if  an  equation  of  the  nth. 
degree  is  in  the  general  form, 
4  If  the  second  term  is  wanting,  the  sum  of  the  roots  is  0. 


THEORY  OF   EQUATIONS  607 

;f  the  last  term  is  wanting,  at  least  one  root  is  0. 
^  We  also  see  that  if  the  last  term  is  integral,  it  is  divisible 
by  every  integral  root. 

nT  720.  If  all  but  one  of  the  roots  of  an  equation  of  the  nth 
degree  in  the  general  form  are  known,  the  remaining  root  may 
be  found  by  changing  the  sign  of  the  coefiicient  of  the  second 
term  of  the  given  equation,  and  subtracting  the  sum  of  the 
known  roots  from  the  result ;  or,  by  dividing  the  last  term  of 
the  given  equation  if  n  is  even,  or  its  negative  if  n  is  odd,  by 
the  product  of  the  known  roots. 

If  all  but  two  are  known,  the  coefficient  of  the  second  term 
of  the  depressed  equation  (§  716)  may  be  found  by  adding  the 
sum  of  the  known  roots  to  the  coefficient  of  the  second  term 
of  the  given  equation;  and  the  last  term  of  the  depressed 
equation  may  be  found  by  dividing  the  last  term  of  the  given 
equation  by  plus  or  minus  the  product  of  the  known  roots 
according  as  n  is  even  or  odd. 

Ex.   Two  roots  of  the*  equation  9  aj^  —  37  a;^  —  8  a;  +  20  =  0 

are  2  and  —  -;  what  are  the  others  ? 
t) 

We  first  put  the  equation  in  the  general  form  by  dividing 

each  term  by  9. 

It  then  becomes  ^*  —  -^  ^^  —  q  ^  +  -^  =  0. 
j7  y  «7 

Since  there  is  no  ic^  term,  the  coefficient  of  the  second  term 
isO. 

Then   the   coefficient  of  the  second  term  of  the  depressed 

5         1 

equation  is  0  +  2  —  - ,  or  -  • 
o  o 

The  coefficient  of  the  last  term  of  the  depressed  equation 

.    20      /     10\  2 

The  depressed  equation  isa^-f-a;  —  -  =  0. 

Solving,  a?  =  Q  or  —  1. 
o 


/ 


J 


508      ADVANCED  COURSE  IN  ALGEBRA 

EXERCISE  119 

1,   Two  roots  of  16  x^  -  37  x  -  21  =  0  are  -  1  and  - ;  find  the  other 

4 
root. 

•      2.  Three  roots  of  4  a;*  -  17  x^  -  64  x'^  +  257  ic  -  60  =  0  are  3,  -  4,  and 

6 ;  find  the  other  root. 

3.  Four  roots  of  2  ac^  _|.  3  ^4  _  30  x^  -  65  a;2  +  18  ic  +  72  =  0  are  1,  -  2, 

Q 

4,  and  — :  find  the  other. 
'  2' 

4.  Five  roots  of    36  x^  -  475  x^  -  431  cc^  +  91  x^  +  71  cc  -  12  =  0    are 

—  1,  —  3,  — ,  -,  and  - ;  find  the  other. 

''236 

5.  One  root  of  x^  +  8  x2  -  23  x  -  210  =  0  is  -  6  ;  find  the  others. 

6.  Two  roots  of  6  x*  +  7  x^  -  37  x'^  -  8  x  +  12  =  0  are  2  and  -;  find 
the  others. 

7.  Three  roots  of  x^  -  6  x*  -  27  x^  +  148  x^  +  204  x  -  720  =  0  are  2, 

—  3,  and  —  4  ;  find  the  others, 

8.  Two  roots  of  x*  -  (2  a2  +  5)  x^  +  6  ax  +  a*  _  5  ^2  +  4  =  0  are  a  -  1 
and  —  a  +  2  ;  find  the  others. 

721.    Symmetrical  Functions  of  the  Roots. 

The  expressions  for  pi,  P2,  "-jPni  iii  §  ''^18,  are  symmetrical 
(§  146)  functions  of  the  roots  of  the  equation,  and  can  be 
expressed  as  follows  (compare  §  150) : 

Pi=  ~  ^a. 

P2  =  Sa6. 

^3  ==  —  "^abc,  etc. 

Other  symmetrical  functions  of  the  roots  can  be  expressed  in 
terms  of  the  coefficients. 

Ex.   If  a,  h,  and  c  are  the  roots  of  the  equation 

find  the  values  of  2a^,  and  ^a-b. 

We  have  ^a'^o?  ^W  -\-  (?={a^-b  -^cf  -2(ah  -{-hc  +  ca) 

=  CXay  -2{^ab)  =  {-p,y  -2p2=p,'-2p2, 
Again,     %a^b  =  a^b  +  a^c  +  b^c  +  b'a  +  c^a  +  c^b 

=  {a  4- b -\- c)  (ab -\- be -{-  cd)  —  3  abc 

=  {-Pi)P2  +  Sp,  =  3ps  -PiP2- 


^ 


THEORY  OF   EQUATIONS  609 


EXERCISE  120 


If  a,  6,  and  c  are  the  roots  of  7?  -^-pxx^  +P2X  +^3  =  0,  find 

1.    si.  2.    S— .  3.    2a3.  4.    Za^\  5.    S— . 

a  ab  o2 

If  a,  6,  c,  and  d  are  the  roots  of  x*  +  piic^  4-  pix"^  +  p^x  +p^  =  0,  find 

6.   si.  7.   S— .  8.    Sa^fec.  9.    2^2;,.  10.    Sa8. 

a  ao  ,  > 

[p^  722.   Fractional  Roots. 

An  equation  in  the  general  form  with  integral  coefficients 
cannot  have  as  a  root  a  rational  fraction  (§  198)  in  its  loivest 
terms. 

Let  the  equation  be 

x^  +^91^-1  +i)2aj"-2  ^  ...  ^p^^x  +p^  =  0, 

where  pi,  p2,  •••,  Pn  are  integral. 

If  possible,  let  -,  a  rational  fraction  in  its  lowest  terms,  be 
b 

a  root  of  the  equation ;  then, 

.   Multiplying  each  term  by  6""^  and  transposing, 

^=.-(p^a--^  -^p,a^n  +  ...  +Pn-iab--'  -\-pJ>''-')- 

By  hypothesis,  a  and  b  have  no  common  divisor;  hence  a" 
and  b  have  no  common  divisor  (§  673). 

We  then  have  a  rational  fraction  in  its  lowest  terms  equal 
to  an  integral  expression,  which  is  impossible. 

Therefore,  the  equation  cannot  have  as  a  root  a  rational 
fraction  in  its  lowest  terms. 

723.   Complex  Roots. 

If  a  complex  number  is  a  root  of  an  equation  in  the  general 
form,  with  real  coefficients,  its  conjugate  (§  425)  is  also  a  root. 

Let  the  equation  be 

a;»-f  pia;"-^+  ...  -\-pn-iX+p^  =  0,  (1) 

where  pi,  «..,  ^\  are  real  numbers. 


510  ADVANCED  COURSE   IN   ALGEBRA 

Let  a  -f  hi,  where  a  and  h  have  the  same  meanings  as  in 
§  415,  be  a  root  of  the  equation ;  then, 

(a  4-  uy  +pla  +  Uy-^  +  •••  +P«-i(«  +  &0+Pn  =  0. 

Expanding  by  the  Binomial  Theorem,  we  have  by  §  411, 

+Pi[a"-^  +  (w  -  l)a^-^U  -  (^-l)(^-^)t^n-352 -| 

+  -4-Pn-i(ot  +  &0+i>-=0-  (2) 

Collecting  the  real  and  imaginary  terms,  we  have  a  result  of 

the  form  P+Qi  =  0.  (3) 

Here,  P  stands  for  the  sum  of  all  the  terms  containing  a 
alone,  together  with  all  the  terms  containing  even  powers  of  t  j 
Qi  for  all  terms  containing  odd  powers  of  i. 

In  order  that  equation  (3)  may  hold,  we  must  have 
P=0,  and  Q  =  0(§418). 

Now  substituting  a  —  hi  for  x  in  the  first  member  of  (1),  it 
becomes 

(a  _  uy  +p,{a  -  Uy-^  +  -  +P„-i(a  -  hi)  -\-p^.         (4) 

Expanding  the  powers  of  a  —  hi,  we  shall  have  a  result  which 
differs  from  the  first  member  of  (2)  only  in  having  the  even 
terms  in  each  expansion,  or  those  involving  i  as  a  factor, 
changed  in  sign. 

Then,  collecting  the  real  and  imaginary  terms,  the  expression 
(4)  equals  P-Qi] 

where  P  and  Q  have  the  same  meanings  as  before. 
But  since  P=0  and  Q  =  0,  P-qi  =  0. 
Hence,  a  —  hi  is  a  root  of  (1). 

The  above  demonstration  holds  without  change  when  a  equals  zero ; 
thus  the  theorem  holds  for  any  pure  imaginary  number  (§  418). 

724.  It  follows  from  §§  713  and  723  that  every  equation  of 
odd  degree  has  at  least  one  real  root;  for  an  equation  cannot 
have  an  odd  number  of  complex  roots. 


THEORY  OF   EQUATIONS  511, 

725.  The  product  of  the  factors  of  the  first  member  of  (1), 
§  723,  corresponding  to  the  conjugate  complex  roots  a-\-b  V—  1 
and  a  — 6  V— 1,  is 

[x-(a 4-  6  V^n:)][a;  -  (a-  6  V^] 
=  (x-  af--{b^'^iy  =  ix  -  af  +  h''', 
and  is  therefore  positive  for  every  real  value  of  x. 

TRANSFORMATION  OF  EQUATIONS 

726.  To  transform  an  equation  into  another  which  shall  have  . 
the  same  roots  with  contrary  signs. 

Let  the  equation  be 

X^  +i>l«^"'  +P2«""'  +  —  +Pn-lX  +Pn  =  0.  (1) 

Substituting  —  ?/  for  x,  we  have 

{-yY+Pi(-yr-'+P2(-yT-'-h-"+Pn-i(-y)-^Pn  =  o. 

Dividing  each  term  by  (—  1)",  we  have 

Or,  y--p^^f-l^p^y-^ ±Pn-i2/Ti?n  =  0;  (2) 

the  upper  or  lower  signs  being  taken  according  as  n  is  odd  or 
even.  -  -  --   -  -- 

>  \      It  follows  from  (1)  and  (2)  that  the  desired  transformation 


may  be  effected  by  changing  the  signs  of  the  terms  of  odd  degree, 
if  the  degree  of  the  equation  is  even,  or  the  signs  of  the  terms  of  ^ 


even  degree  and  of  the   independent  term^  if  the  degree  of  the 
'  equation  is  odd.  v'*^     ' "''  ^ '  -    ^^   ^     '  '   "^ 

The  above  rule  applies  whether  the  equation  is  complete  or  incomplete. 

Ex.  Transform   the  equation  x^  —  10x-\-4:  =  0  into  another 
which  shall  have  the  same  roots  with  contrary  signs. 

By  the  rule,  the  transformed  equation  is  a^  —  10  a;  —  4  =  0. 

727.    To  transform  an  equation  into  another  whose  roots  shall 
be  respectively  m  times  those  of  the  first. 
Let  the  equation  be 

a;"  +  2)iX"-^  +  p2a;"-2  -\ \-  p,^_^x  4-  Pn  =  0. 


o 


512  ADVANCED  COURSE  IN  ALGEBRA 

Putting  mx  =  y,  that  is,  ^  for  x,  we  have 
m 

Multiplying  each  term  by  m**, 

2/"  +i)im2/^~'  +i?2my-'  +  ••  •  +i>«_im"-^2/  +Pnm''  =  0. 
^  jij  Hence,  to   effect   the   desired   transformation,   multiply   the 
*^      ijiecond  term  by  m,  the  third  term  by  m^,  and  so  on. 

Ex.  Transform  the  equation  £c^  +  7ic^  —  6  =  0  into  another 
whose  roots  shall  be  respectively  4  times  those  of  the  first. 

Supplying  the  missing  term  with  the  coefhcient  zero,  and 
applying  the  rule,  we  have 

x3  +  4.7ar  +  42.0ic-43.6  =  0,  ora53_|.28a^-384  =  0. 

^M  728.  To  trayisform  an  equation  with  fractional  coefficients  into 
another  whose  coefficients  shall  be  integral,  that  of  the  first  term 
being  unity. 

The  transformation  may  be  effected  by  transforming  the 
equation  into  another  whose  roots  shall  be  respectively  m  times 
those  of  the  first  (§  727),  and  then  giving  m  such  a  value  as 
will  make  every  coefficient  integral. 

By  giving  to  m  the  least  value  which  will  make  every  coeffi- 
cient integral,  the  result  will  be  obtained  in  its  simplest  form. 

Ex.    Transform  the  equation  oi? —-\ =  0  into  an- 

3      36     108 
other  whose  coefficients  shall  be  integral,  that  of  the  first  term 
being  unity. 

By  §  727,  the  equation 

3         36        108 
has  its  roots  respectively  m  times  those  of  the  given  eqiiation. 
It  is  evident,  by  inspection,  that  the  least  value  of  m  which 
will  make  every  coefficient  integral,  is  6. 
Putting  m  =  Qj  we  have 

ix?-2x^-x  +  2  =  0, 
whose  roots  are  6  times  those  of  the  given  equation. 


THEOllY   OF   EQUATIONS  518 

i^^'    729.    To  transform    an   equation    into   another  whose  roots 
f  shall  he  respectively  those  of  the  first  increased  by  m.  '^^.  ^  7>\  -fff^ 

Let  the  equation  be 

a;»+i>ia^rV-i:-  +  Pn-i^  +Pn  =  0.  (1) 

Putting  x-\-m  —  y,  that  is,  y  —  m  for  x,  we  have 

(?/-m)"+|>i(2/-m)^-i+  ...  +i)n-i(y-m)+p„  =  0.      (2) 

Expanding  the  powers  oi  y  —  m  hj  the  Binomial  Theorem, 
and  collecting  the  terms  involving  like  powers  of  y,  we  shall 
have  a  result  of  the  form 

rj.  q,y--'  +  -  +  Qn-iV  +  gn  =  0,  (3) 

whose  roots  are  respectively  those  of  the  given  equation  in- 
creased by  m. 

Ex.  Transform  the  equation  a^  —  7cc  +  6  =  0  into  another 
whose  roots  shall  be  respectively  those  of  the  first  increased 
by  2. 

Substituting  i/  —  2  for  x, 

(2/-2)3_7(2/-2)+6  =  0. 

Expanding,  and  collecting  the  terms  involving  like  powers 
of  2/,  we  have  f  _Qy2  _^^y  ^i2  =  0. 

730.  If  m  and  the  coefficients  of  the  given  equation  are 
integral,  the  coefficients  of  the  transformed  equation  may  be 
conveniently  found  by  the  following  method. 

Putting  x-\-m  for  y  in  (3),  we  obtain 

(x^rny^q^(x-{-m)^-^+  ...  +  g._i  (x -f  ??i)  +  g„  =  0 ;     (4) 

which  must,  of  course,  take  the  same  form  as  (1)  on  expanding 

the  powers  of  x  +  m,  and  collecting  the  terms  involving  like 

powers  of  x. 

Dividing  the  first  member  of  (4)  by  a;  +  m,  we  have 
-       (x  +  7ny-^  +  q,(x  +  my-^-{-  -••  -\-qn-^(x-{-m) -^q^-i      (6) 

as  a  quotient,  with  a  remainder  q^. 

Dividing  (5)  hj  x-{-  m,  we  have  the  remainder  q^-i ;  etc. 


II 


514      ADVANCED  COURSE  IN  ALGEBRA 

Hence,  to  obtain  the  coefficients  of  the  transformed  equation  : 

Divide  the  first  member  of  the  given  equation  by  x  +  m;  the 
remainder  will  be  the  last  term  of  the  required  equation. 

Divide  the  quotient  just  found  by  x-\-m;  the  remainder  will  be 
the  coefficient  of  the  next  to  the  last  term  of  the  transformed  equa- 
tion; and  so  on. 

Ex.  Transform  the  equation  a^  —  7a;  +  6  =  0  into  another 
whose  roots  shall  be  respectively  those  of  the  first  increased 
by  2. 

Dividing  a;^  —  7  a;  -f-  6  by  aj  +  2,  we  have  the  quotient  a^  —  2  a; 
—  3,  and  the  remainder  12. 

Dividing  x^  —  2  x  —  ^  by  x-{-2,  we  have  the  quotient  x  —  4, 
and  the  remainder  5. 

Dividing  a;  —  4  by  a;  +  2,  we  have  the  remainder  —  6. 

Then,  the  transformed  equation  is 

i^_6a;2_^5aj  +  12  =  0. 
Compare  Ex.,  §  729. 

731.  To  transform  an  equation  into  another  whose  roots 
shall  be  those  of  the  first  diminished  by  m,  we  change  y  —  mto 
y  -f  m  in  the  method  of  §  729,  and  x-\-m  to  x  —  m  in  the  rule 
of  §  730. 

732.  To  transform  the  equation 

where  2h  'is  not  zero,  into  another  whose  second  term  shall  be 
wanting. 

Expanding  the  powers  of  y  —  mm  the  first  member  of  (2), 
§  729,  and  collecting  the  terms  involving  like  powers  of  y,  we 

y""  -\-{2\—mn)y'^    +  •••  =0. 

If  m  be  so  taken  that  pi  —  mn  —  0,  whence  m  =  — ,  the  coeffi- 
cient of  ?/""^  will  be  zero. 

Hence,  the  desired  transformation  may  be  effected  by  sub- 
stituting in  the  given  equation  y  —  ~  in  place  of  x. 


THEORY  OF   EQUATIONS  616 

733.   Synthetic  Division. 

The  operation  of  division,  in  examples  like  that  of  §  730,  may 
be  conveniently  performed  by  a  process  known  as  Synthetic 
Division. 

Let  it  be  required  to  divide  a^  —  12  x-  +  29  x  —  21  by  a;  —  3. 

Using  detached  coefficients  (§  104),  we  have 


1  _  12  +  29  -  21 

1-3 

1-3 

1-9  +  2, 

-    9 

-    9  +  27 

+   2 

+    2-6 

-15, 

Remainder. 

We  may  omit  the  first  term  of  each  partial  product,  for  it  is 
merely  a  repetition  of  the  term  immediately  above. 

Also,  the  second  term  of  each  partial  product  may  be  added 
to  the  corresponding  term  of  the  dividend,  provided  we  change 
the  sign  of  the  second  term  of  the  divisor  before  multiplying. 

The  work  now  stands  : 


1-12  +  29-21 
+    3 
-    9 

-27 
+    2 

+    6 


1+3 


1-9  +  2 


-15 

The  first  term  of  the  divisor  being  unity  in  all  applications 
of  §  730,  it  may  be  omitted ;  and  the  first  terms  of  the  succes- 
sive dividends  constitute  the  quotient. 

Raising  the  oblique  columns,  the  operation  will  stand  as 
follows : 

Dividend,  1    -12    +29   -21  |  +3 

Partial  products,         _    +    3    —27    +    6 

Quotient,  1    —    9    +    2,-15  Remainder. 

The  complete  result  is  obtained  as  follows : 


616       ADVANCED  COURSE  IN  ALGEBRA 

Multiplying  the  first  term  of  the  dividend  by  3,  and  adding 
the  result  to  the  second  term  of  the  dividend,  gives  the  second 
term  of  the  quotient. 

Multiplying  the  latter  by  3,  and  adding  the  result  to  the 
third  term  of  the  dividend,  gives  the  last  term  of  the  quotient. 

Multiplying  the  latter  by  3,  and  adding  the  result  to  the  last 
term  of  the  dividend,  gives  the  remainder. 

Hence,  the  quotient  is  x^  —  9x-\-2,  and  the  remainder  — 15. 

If  the  term  involving  any  power  of  x  is  wanting,  it  must  be  supplied 
with  the  coefficient  0  before  applying  the  rule. 

The  work  of  transforming,  by  Synthetic  Division,  the 
equation  ^_7a;4.6  =  0 

into  another  whose  roots  shall  be  respectively  those  of  the  first 
increased  by  2,  will  stand  as  follows  (compare  §  730) : 

1  +0-7  +    6|  -2 
-2  +4  +    6 

-  2  -  3  +12,  1st  Rem. 
-2  +8 

-  4  +  5,  2d  Rem. 
-2 

-  6,  3d  Rem. 

Thus,  the  transformed  equation  is 

a^_6a^  +  5i»  +  12  =  0. 

EXERCISE  121 

Transform  each  of  the  following  into  an  equation  which  shall  have  the 
same  roots  with  contrary  signs : 

1.    x*- 8x3 -7x2  + 3x +  4  =  0.  2.    x*^  +  6x*  -  2x  -  5  =  0. 

3.  Transform  x^  +  10  x^  +  5  x  —  7  =  0  into  an  equation  whose  roots 
shall  be,  respectively,  5  times  those  of  the  first. 

4.  Transform  x*  —  4  x^  +  2  x^  +  3  =  0  into  an  equation  whose  roots 
shall  be,  respectively,  —  6  times  those  of  the  first. 

6.   Transform  Sx^  +  Sx^  —  2  =  0  into  an  equation  whose  roots  shall 
3 
be,  respectively,  -  those  of  the  first. 


THEORY  OF  EQUATIONS  617 

6.   Transform  7ic*  +  6x3  —  75a;  +  125  =  0  into  an  equation  whose 

roots  shall  be,  respectively,  those  of  the  first  multiplied  by 

5 

Transform  each  of  the  following  into  an  equation  with  integral  coeflB- 

cients,  that  of  the  first  term  being  unity : 

4        32  18         54   ^  72 

8.   xB  +  ^-ll  =  0.  10.   :^+^  +  l^-J-  =  0. 

25      40  5        125      225 

11.  Transform  x^  —  5x^  -{■  6x  +  11  =0  into  an  equation  whose  roots 
shall  be,  respectively,  those  of  the  first  diminished  by  5. 

12.  Transform  x^  —  4:x'^  —  3x  —  29  =  0  into  an  equation  whose  roots 
shall  be,  respectively,  those  of  the  first  increased  by  6. 

13.  Transform  x^  +  13  x^  _  §2  =  0  into  an  equation  whose  roots  shall 
be,  respectively,  those  of  the  first  increased  by  2. 

14.  Transform  x^  +  2  x^  +  x^  —  7  x  +  31  =  0  into  an  equation  whose 
roots  shall  be,  respectively,  those  of  the  first  diminished  by  1. 

15.  Transform  2c*  —  3  ic^  +  8  cc^  _  iq  =  0  into  an  equation  whose  roots 
shall  be,  respectively,  those  of  the  first  increased  by  3. 

16.  Transform  x^  +  Qx"^  +  6x  +  19  =  0  into  an  equation  whose  roots 
shall  be,  respectively,  those  of  the  first  diminished  by  4. 

0r}(^  DESCARTES'  RULE  OF  SIGNS 

734.  If  an  equation  of  the  nth  degree  is  in  the  general 
form  (§  712),  a  Permanence  of  sign  occurs  when  two  succes- 
sive terms  have  the  same  sign,  and  a  Variation  of  sign  occurs 
when  two  successive  terms  have  opposite  signs. 

Thus,  in  the  equation  a;^  —  3aj*  —  a^-|-5ic-|-l=0,  there  are 
two  permanences  and  two  variations. 

735.  Descartes'  Rule  of  Signs. 

No  equation  J  tchether  complete  or  incomplete,  can  Jmve  a 
greater  number  of  positive  roots  tha7i  it  has  variations  of  sign; 
and  no  complete  equation  can  have  a  greater  number  of  nega- 
tive roots  than  it  has  permanences  of  sign. 

Let  an  equation  in  the  general  form  have  the  following 

signs:  +    +  0   _   +  0  0 , 

the  missing  terms  being  supplied  with  zero  coefficients. 


+ 

+ 

0    - 
-    0 

+ 

+ 

0  0 
-    0 

0    4- 

+ 

+ 
1 

m 

2 

3     4 

+ 
5 

-    0 

6     7 

—    m 
8     9 

4- 
10 

518  ADVANCED   COURSE   IN  ALGEBRA 

If  we  introduce  a  new  positive  root  a,  we  multiply  this  by 
aj  —  a  (§  717) ;  writing  only  the  sig7is  which  occur  in  the 
process,  we  have 

12    3    456789 

4-+0-+00--  (1) 

+   ~ 


(2) 


where  m  signifies  a  term  which  may  be  +,  0,  or  — . 
.    Now,  in  (1),  let  a  dot  be  placed  over  the  first  minus  sign, 
then  over  the  next  plus  sign,  then  over  the  next  minus  sign, 
and  so  on. 

The  number  of  dots  shows  the  number  of  variations ;  thus, 
in  (1)  there  are  three  variations. 

In  the  above  result,  we  observe  the  following  laws : 

I.  Directly  under  each  dotted  term  of  (1)  is  a  term  of  (2) 
having  the  same  sign. 

Thus,  the  terms  numbered  4,  5,  and  8,  in  (1)  and  (2),  have 
the  same  sign. 

II.  The  last  term  of  (2)  is  of  opposite  sign  to  the  term 
directly  under  the  last  dotted  term  of  (1). 

The  above  laws  are  easily  seen  to  hold  universally. 

By  the  first  law,  however  the  term  marked  m  is  taken, 
there  are  at  least  as  many  variations  in  the  first  eight  terms  of 
(2)  as  in  (1) ;  and  by  the  second  law,  there  is  at  least  one 
variation  in  the  remaining  terms  of  (2). 

Hence,  the  introduction  of  a  new  positive  root  increases  the 
number  of  variations  in  the  equation  by  at  least  one. 

If,  then,  we  form  the  product  of  all  the  factors  correspond- 
ing to  the  negative  and  imaginary  roots  of  an  equation,  multi- 
plying the  result  by  the  factor  corresponding  to  each  positive 
root  introduces  at  least  one  variation. 

Hence,  the  equation  cannot  have  a  greater  number  of  posi- 
tive roots  than  it  has  variations  of  sign. 


<«-6frw^4-  ^     t^^^    ^^f^   f>a^^  .J'>^'  ^  l^*^*- 

-v«>./7  f^  .  A^5  f  liEORY  OF  EQUATIONS  ' '        '    519  ^ 

To  prove  the  second  part  of  Descartes'  Rule,  let  —  ?/  be 
substituted  for  x  in  any  complete  equation. 

Then  since  the  signs  of  the  alternate  terms  commencing  with 
the  second  are  changed  (§  726),  the  original  permanences  of 
sign  become  variations. 

But  the  transformed  equation  cannot  have  a  greater  number 
of  positive  roots  than  it  has  variations. 

Hence,  the  original  equation  cannot  have  a  greater  number 
of  negative  roots  than  it  has  permanences. 

In  all  applications  of  Descartes'  Kule,  the  equation  must  contain  a 
term  independent  of  x ;  that  is,  no  root  must  equal  zero  (§  182)  j  for  a 
zero  root  cannot  be  regarded  as  either  positive  or  negative. 

"y  736.  It  follows  from  the  last  part  of  §  735,  and  from  §  726, 
that  in  any  equation,  complete  or  incomplete,  the  number  of 
negative  roots  cannot  exceed  the  number  of  variations  in  the 
equation  which  is  formed  from  the  given  equation  by  changing 
the  signs  of  the  terms  of  odd  degree. 

^r^l2n.   In  any  complete  equation,  the  sum  of  the  number  of 

permanences  and  variations  equals  the  number  of  terms  less 

one,  or  the  degree  of  the  equation. 

That  is,  the  sum  of  the  number  of  permanences  and  variar 

tions  equals  the  number  of  roots  (§  715). 

\    Hence,  if  the  roots  of  a  complete   equation  are  all  real, 
1   the  number  of  positive  roots  equals  the  number  of  variations,   '"^ 

and  the  number  of  negative  roots  equals  the  number  of  per- 
'  manences. 

An  equation  whose  terms  are  all  positive  can  have  no  posi-  )  w 

tive  root ;  and  a  complete  equation  whose  terms  are  alternately   / 

positive  and  negative  can  have  no  negative  root. 

738.  Ex.   Determine  the  nature  of  the  roots  of 
a^  +  2a;  +  5  =  0. 

There  is  no  variation,  and  consequently  no  positive  root. 
Changing  the  sign  of  the  independent  term,  we  have 

a^  +  2a;-5  =  0. 


520      ADVANCED  COURSE  IN  ALGEBRA 

Here  there  is  one  variation ;  and  therefore  the  given  equa- 
tion cannot  have  more  than  one  negative  root  (§  736). 

Then  since  the  equation  has  three  roots  (§  715),  one  of  them 
must  be  negative  and  the  other  two  imaginary. 

If  two  or  more  successive  terms  of  an  equation  are  wanting,  it  follows 
by  Descartes'  Rule  that  the  equation  must  have  imaginary  roots. 

EXERCISE   122 

If  the  roots  of  the  following  are  all  real,  determine  their  signs : 

1.  x^  +  x'^-Ux-U4:  =  0.         3.   x*-a;3-19a;2  +  49x-30  =  0. 

2.  4x^-2Zx2  +  Ux  +  6  =  0.     4.  5ic4-43x3^112ic2-68x-48  =  0. 

5.  x*  -  4  x^  -  23  x2  +  54  .r  +  72  =  0. 

6.  x5-lla;*  +  33a:3+llx2-154x-120  =  0. 

7.  2  a;5  +  29x*  + 119x3  + 159x2  + 7  X -60  =  0. 

Determine  the  nature  of  the  roots  of  the  following: 

8.  x3- 2x2- 3  =  0.  11.   x5  +  4x3-l=0. 

9.  2x*  +  5x2  +  4  =  0.  12.   3x6-5  =  0. 

10.    x5  +  32=0.  13.    x7  + 3x4 +  5x2+ 2  =  0. 

14.  Prove  that  the  equation  x^  +  x^  —  x2  +  3  =  0  has  at  least  two  imagi- 
nary roots. 

LIMITS  TO  THE  ROOTS 

^■T*^      739.    To  find  a  superior  limit  to   the  positive  roots  of  an 
equation. 

The  following  examples  illustrate  the  process  of  finding  a 
superior  "limit  to  the  positive  roots  of  an  equation. 

1.   Find  a  superior  limit  to  the  positive  roots  of 

Grouping  the  positive  and  negative  terms,  we  can  write  the 
first  member  in  the  form 

a?{x-Z)  +  2{x~^.  (1) 

It  is  evident  that  if  x  equals  or  exceeds  3,  the  expression  (1) 
is  positive. 


THEORY   OF   EQUATIONS  521 

Hence,  no  root  of  the  given  equation  equals  or  exceeds  3, 
and  3  is  a  superior  limit  to  the  positive  roots. 

2.   Find  a  superior  limit  to  the  positive  roots  of 

x^  - 15  x'  - 10  a.'  +  24  =  0. 

2  x^  X* 

We  separate  the  first  term  into  the  parts and  — ,  and 

write  the  first  member  in  the  form 

f^^l5A  +  (^-10x\+24.,ov'^(2x'-4.5)  +  ^(x^-^^^ 

It  is  evident  from  this  that  no  root  can  be  so  great  as  5} 
hence,  5  is  a  superior  limit  to  the  positive  roots. 
■  If  we  had  written  the  first  member  in  the  form 

(t  _  15  xA  +  (-  -  10  xW  24,  or  ^(x^  -  30)  +  -(x^-  20)  +  24, 

we  should  have  found  6  as  a  superior  limit  to  the  positive  roots. 

Thus,   separating  x*  into  -^  and  ^,   instead  of  ^  and  — ,  gave  a 
smaller  limit. 


.Ji 


740.  To  find  an  mferior  limit  to  the  negative  roots  of  an 
equation. 

First  transform  the  equation  into  another  which  shall  have 
the  same  roots  with  contrary  signs  (§  726). 

The  superior  limit  to  the  positive  roots  of  the  transformed 
equation,  obtained  as  in  §  739,  with  its  sign  changed,  will  be 
an  inferior  limit  to  the  negative  roots  of  the  given  equation. 

Ex.   Find  an  inferior  limit  to  the  negative  roots  of 

a^  +  2ar^H-5a^-7  =  0. 
Changing  the  signs  of  the  x^  term  and  the  independent  term 
(§  726),  we  have       a^  +  2a^-5a^  +  7  =  0.  (1) 

We 'can  write  the  first  member  in  the  form 

It  is  evident  from  this  that  no  root  of  (1)  can  be  so  great  as 
2;. hence,  —  2  is  an  inferior  limit  to  the  negative  roots  of  tho 
given  equation. 


522       ADVANCED  COURSE  IN  ALGEBRA 

By  grouping  the  x^  and  x"^  terms,  in  (1),  we  obtain  a  smaller  limit  than 
if  we  group  tlie  x^  and  x:^  terms. 

EXERCISE  123 

In  each  of  the  following,  find  a  superior  limit  to  the  positive  roots,  and 
an  inferior  limit  to  the  negative  : 

1.  x3  +  3x2  +  x-4  =  0.  4.  3x4 -5x2 -8a:- 7  =  0. 

2.  x4  +  5x3-15x-9  =  0.      5.  x^  -  4x*  + 6x3  + 32x2  -  15x+3  =  0. 

3.  x4  +  3x2-5x-8  =  0.        6.   2x5  +  5x*  +  6x3  -  13x2 -25x  +  4  =  0. 

7.  In  the  equation  x^- 2x2-^  3x+ 1  =0,  prove  3  a  superior  limit  to 
the  positive  roots,  and  -  2  an  inferior  limit  to  the  negative. 

8.  In  the  equation  2  x^  +  5 x2  -  7  x  -  3=  0,  prove  -  4  an  inferior  limit 
to  the  negative  roots,  and  find  a  superior  limit  to  the  positive. 

9.  In  the  equation  x>  +  3  x^  -  9  x2  +  12  x  -  10  =  0,  prove  3  a  superior 
limit  to  the  positive  roots,  and  -  6  an  inferior  limit  to  the  negative. 

LOCATION  OF  ROOTS  ^f^  ? 

/tj^    741.    If  two  real  yiumhers,  a  and  b,  not  roots  of  the  equation 

f(x)  =  x^  4-Piaj"-'  +  •  •  •  +  Pn-lX  +i>n  =  0, 

ivhen  substituted  for  x  in  f(x),  give  results  of  opposite  sign,  an 
odd.  number  of  roots  off(x)  =  0  lie  between  a  and  b. 

Let  a  be  algebraically  greater  than  b. 

Let  d,  '"y  ghQ  the  real  roots  of /(«)  =  0  lying  between  a  and 
b,  and  h,  "•,  k  the  remaining  real  roots. 

Then,  by  §  717, 

f(x)=^{x-d)-'{x-g)-(x-h).-.(x-k)-F{x)',         (1) 

where  F(x)  is  the  product  of  the  factors  corresponding  to  the 
complex  roots  off(x)  =0. 

Substituting  a  and  b  for  x  in  (1),  we  have  (§  251), 

f(a)  =  (a-d)  ...  (a-g)  -  (a-h)  ...  {a-h)'F(a\    . 
.       and        f(b)  =  {b-d)  ...  (fi-g)  -  (b-h)  ...  {b-7c)'F(b). 

Since  each  of  the  numbers  c?,  ...,(/  is  less  than  a  and  greater 
than  b,  each  of  the  factors  a  —  d,-",a  —  gis  positive,  and  each 
of  the  factors  b  —d,  •••,  b  —  g  negative. 


THEORY  OF  EQUATIONS  523 

Again,  since  none  of  the  numbers  h,  •••,  k  lie  between  a  and 
h,  the  expression  (a  —  li)  •  •  •  (a  —  k)  has  the  same  sign  as  (h  —  h) 

■■■{b-k). 

Also,  F(a)  and  F(p)  are  positive;  for  the  product  of  the 
factors  corresponding  to  a  pair  of  conjugate  complex  roots 
is  positive  for  every  real  value  of  x  (§  725). 

But  by  hypothesis,  /(a)  and  f(h)  are  of  opposite  sign. 

Hence,  the  number  of  factors  b  —  d,  •••,  b  —  g  must  be  odd  ; 
that  is,  an  odd  number  of  roots  lie  between  a  and  b. 

If  the  numbers  substituted  differ  by  unity,  it  is  evident  that  the  in- 
tegral part  of  at  least  one  root  is  known. 

Ex.     Locate  the  roots  of  a^  +  a^  —  6ic  —  7  =  0. 

By  Descartes'  Rule  (§  735),  the  equation  cannot  have  more 
than  one  positive,  nor  more  than  two  negative  roots. 

The  values  of  the  first  member  for  the  values  0,  1,  2,  3,-1, 
—  2,  and  —  3  of  a;  are  as  follows : 

x  =  0',    -7.       x  =  2',    -7.       x=-l',    -1.       x=-S',    -7. 

x  =  l',    -11.     x==3',    11.         x=-2;   1. 

Since  the  sign  of  the  first  member  is  — when  x  =  2,  and         ' 
H-  when  a;  =  3,  one  root  lies  between  2  and  3.  >^ 

The  others  lie  between  —  1  and  —  2,  and  —  2  and  —  3,  re- 
spectively. 

In  locating  roots  by  the  above  method,  first  make  trial  of  the  numbers 
0,  1,2,  etc.,  continuing  the  process  until  the  number  of  positive  roots  deter- 
mined is  the  same  as  has  been  previously  indicated  by  Descartes'  Rule. 

Thus,  in  the  above  example,  the  equation  cannot  have  more  than  one 
positive  root ;  and  when  one  has  been  found  to  lie  between  2  and  3,  there 
is  no  need  of  trying  4,  or  any  greater  positive  number. 

The  work  may  sometimes  be  abridged  by  finding  a  superior  limit  to 
the  positive  roots,  and  an  inferior  limit  to  the  negative  roots  of  the  given 
equation  (§§  739,  740),  for  no  number  need  be  tried  which  does  not  fall 
between  these  limits. 

EXERCISE  124 

Locate  the  roots  of  the  following  : 

1.  x3  +  4  x2  -  6  =  0. 

2.  ic8  -  7  x2  +  6  X  +  6  =  0. 


524  ADVANCED  COURSE  IN  ALGEBRA 

3.    x*  +  3  x3  -  4  X  -  1  =  0.  4.    x*  +  x3  -  19  x2  -  17  X  +  1  =  0. 

5.  Prove  that  the  equation  x^  -\-2x^  -}-  5x  +  6  =  0  has  either  one  or 
three  roots  between  —  1  and  —  2. 

6.  Prove  that  the  equation  x^  —  5  x^  —  7  x  —  2  =  0  has  a  root  between 
2  and  3,  and  at  least  one  between  0  and  —  1. 

7.  Prove  that  the  equation  x*  —  3x3  +  x2  —  3x  —  4  =  0  has  a  root  be- 
tween 0  and  —  1,  and  at  least  one  between  3  and  4. 

^yiJ^   742.   Location  of  Roots  by  Synthetic  Division. 

With  the  notation  of  §  741,  if  f(a)  and/(&)  are  of  opposite 
sign,  an  odd  number  of  roots  oif(x)  =  0  lie  between  a  and  b. 

Now  by  §  251,  f(a)  =  a-  +p,a^-'  +  •-.  +Pn-ia  +Pn,  (1) 

and  fgj)  =  -b-^p^h--^Jr"'+Pn-,h^-pn.  (2) 

Also,  (1)  and  (2)  are  the  remainders  obtained  by  dividing 

a;"  +  p^x""-^  H h  p^_^x  -\-p^ 

hj  x  —  a  and  x—b,  respectively  (§  139). 

Hence,  if,  when  f(x)  is  divided  by  a?  —  a  and  x—b,  the 
remainders  are  of  opposite  sign,  an  odd  number  of  roots  of 
f(x)  =  0  lie  between  a  and  b. 

The  remainders  may  be  obtained  by  Synthetic  Division. 

Ex.     Locate  the  roots  of  x^  +  x"^  —  5  x  —  A  =  0. 

By  Descartes'  Eule,  the  equation  cannot  have  more  than  one 
positive,  nor  more  than  two  negative  roots. 

Dividing  oi^-\-x-  —  5x  —  4:  by  x,  the  remainder  is  —  4.         (3) 

Dividing  the  first  member  successively  by  ic  —  1,  x  —  2,  x  —  3, 
x-\-l,  x-i'2,  and  x-\-3,  we  have 

1  4_i  _5  _4  Lj.     (4) 

1  2-3 

2  -3  -7 
1  +1   _5  _4  |_2     (5) 

2  6       2 

3  1-2 

1   -!_  1  _  5  _  4  |_3     (6) 

3  12     21 

4  7     17 


+1 
-1 

0 

-5 

0 

-5 

-4|-1 
5 
1 

(7) 

+  1 

-2 
-1 

-5 

2 
-3 

-4|-2 
6 

2 

(8) 

+  1 
-3 

-2 

-5 
6 
1 

-4|-3 

-3 

-7 

(9) 

THEORY  OF   EQUATIONS  525 

In  (5)  and  (6),  the  remainders  are  —  2  and  + 17,  respectively ; 
hence  one  root  lies  between  2  and  3. 

In  (3)  and  (7),  the  remainders  are  —  4  and  + 1,  respectively; 
hence  a  root  lies  between  0  and  —  1. 

In  like  manner,  a  root  lies  between  —  2  and  —  3. 

The  above  process  is  nothing  more  than  a  convenient  way  of  applying 
the  test  of  §  741. 

It  has  moreover  the  advantage  over  the  method  of  direct  substitution 
that,  when  the  integral  part  of  a  root  has  been  found,  the  v^rork  performed 
is  identical  with  the  first  part  of  Horner's  method  (§  794)  for  deter- 
mining additional  root-figures ;  thus,  in  the  above  example,  the  work  in 
(5)  is  identical  with  the  first  three  lines  of  the  determination  by  Horner's 
method  of  the  root  of  the  given  equation  lying  between  2  and  3. 

The  note  to  §  741  applies  with  equal  force  to  the  method  of  §  742. 

EXERCISE  125 

Locate  the  roots  of  the  following  : 

1.  a;3  +  3  x2  -  7  ic  +  2  =  0.  3.  a;*  -  4  a;^  +  6  x  -  2  =  0. 

2.  a;3  +  4  x2  +  ic  -  3  =  0.  4.  x*  -  7  x^  +  x  +  4  =  0. 

6.  Prove  that  the  equation  x^  +  6x  +  'i  =  0  has  one  root  between  0 
and  —  1. 

6.  Prove  that  the  equation  x*  +  2  x^— 5  x^  —  4  x  -  6  =  0  has  a  root  be- 
tween 2  and  3,  and  at  least  one  between  —  3  and  —  4. 

743.  The  methods  of  §§  741  and  742,  though  simple  in 
principle,  and  easy  to  apply,  are  not  sufficient  to  deal  with 
every  problem  in  location  of  roots. 

Let  it  be  required,  for  example,  to  locate  the  roots  of 

By  §  724,  the  equation  has  at  least  one  real  root. 

By  Descartes'  Eule,  it  has  no  positive  root. 

By  §  740,  —  3  is  an  inferior  limit  to  the  negative  roots. 

Putting  X  equal  to  0,  —  1,  —  2,  —  3,  respectively,  the  corre- 
sponding values  of  the  first  member  are  1,  1,  1,  and  —5, 
respectively. 

Then,  the  equation  has  either  one  root  or  three  roots  between 
—2  and  —3;  but  the  methods  already  given  are  not  sufficient 
to  determine  which. 


526 


ADVANCED   COURSE  IN  ALGEBRA 


Sturm's  Theorem  (§  758)  affords  a  method  for  determining 
completely  the  number  and  situation  of  the  real  roots  of  an 
equation. 

It  IS  more  difficult  to  apply  than  the  methods  of  §§  741  and 
742,  and  should  be  used  only  in  cases  which  the  latter  cannot 
resolve. 


-r 


744.  Graphical  Representation. 
The  graph  of  an  expression  of  higher  degree  than  the  second, 

with  one  unknown  number,  may  be  found  as  in  §  465. 

Ex.  Find  the  graph  of 

x^-2x'-2x-{-S. 
Fiity  =  x'-2x^-2x-\-3. 
lix=0,y=3.         Ux=S,      y=6. 
If  x=l, y=0.        If  x=-l,  y=2. 
If  x=2,  y=  -1.     If  x=  -2,  y=-d. 
etc. 

The  graph  is  the  curve  ABC,  which  extends  in  either  direc- 
tion to  an  indefinitely  great  distance  from  XX'. 

745.  Graphical  Location  of  Roots. 

The  principle  of  §  280  holds  for  the  graph  of  the  first  mem- 
ber of  an  equation  of  higher  degree  than  the  second,  with  one 
unknown  number. 

Thus,  the  graph  of  §  744  intersects  XX'  at  oj  =  1,  between 
x  =  2  and  x  =  3,  and  between  x  =  —l  and  x  =  —  2. 

And  the  equation  a;^  —  2a^  —  2aj  +  3  =  0  has  one  root  equal  to 
1,  one  between  2  and  3,  and  one  between  —  1  and  —  2. 

This  may  be  verified  by  solving  the  equation  ;  the  factors  of  the  first 
member  are  x  —  1  and  x'^  —  x  —  S. 

This  method  of  locating  roots  is  simply  a  graphical  represen- 
tation of  the  process  of  §  741,  and  is  subject  to  the  limitations 
stated  in  §  743. 

If  the  graph  does  not  intersect  XX',  the  equation  has  no  real 
root. 


THEORY  OF   EQUATIONS  527 

The  note  to  §  741  applies  with  equal  force  in  the  graphical  method  of 
locating  roots. 

EXERCISE  126 

Locate  the  roots  of  the  following  equations  graphically :  , 

1.  a;3-3a;-l=0.      3.  x^-1  x'^-\-12x-6=0.     5.  x3-8a:2+19x-12=0. 

2.  x*+2ic2+3=0.     4.  x^+7 x'^+Ux+S=0.     6.  x^-Sx'^-2x+6=0. 

7.  xH2x3-6a;2_7a;+6=0. 


^r^^ 


^  T*  DIFFERENTIATION 

746.  Derivatives. 

In  any  function  of  x  (§  250),  let  x-\-h  be  substituted  for  x ; 
subtract  from  the  result  the  given  function,  and  divide  the 
remainder  by  h. 

The  limiting  value  of  the  result  as  h  approaches  the'  limit 
zero,  is  called  the  derivative  of  the  function  with  respect  to  x. 

Let  it  be  required,  for  example,  to  find  the  derivative  of 

jc3  _  2  ^2  +  5 
with  respect  to  x. 

Substituting  x  +  h  for  x,  and  subtracting  from  the  result  the  given 
function,  we  have 

(x  -]-hy-2(x  +  hy  +  6-  (x3  -  2  x2  +  5) 

=  Sx'^h  +  Sxh'^-hh^-4:Xh-2  h?. 

Dividing  this  result  by  h,  we  have 

3  x2  +  3  a;;^  +  7i2  -  4  X  -  2  /i.  (1) 

The  limiting  value  of  (1)  as  h  approaches  0,  is  3  x^  —  4  x. 

Hence,  the  derivative  oi  x^  —  2  x'^  ■\-  b  with  respect  to  a;  is  3  x"  —  4  «. 

The  process  exemplified  above  is  called  Differentiation. 

747.  In  general,  let  u  represent  any  function  of  x;  and 
suppose  that,  when  x  is  changed  to  a;  +  ^,  u  becomes  u  +  h\ 

Then,  the  derivative  of  u  with  respect  to  x  is  expressed  as 
follows  (compare  §  265), 


lira  V{u  +  h^  —  u\ 


528       ADVANCED  COURSE  IN  ALGEBRA 

It  follows  from  the  above  that  —u=  ,^^!^^ [-1;  (1) 

dx        ^^  =  0^ 

d  . 

where  —  u  stands  for  the  derivative  of  u  with  respect  to  x. 
dx 

748.  The  process  of  differentiation  is  facilitated  by  means 
of  the  following  formulse,  in  which  a  represents  any  constant, 
n  any  positive  integer,  and  w,  v,  w,  •••,  any  functions  of  x : 

I.   ^x  =  l. 
dx 

11.   ±(u-\-a)=^u. 
dx  dx 

III.   —(au)  =  a  —  u. 
dx  dx 

dx  dx        dx        dx 

V.     (UVW  '••)  =  (VW  •••) U-{-(uW  •••) V+"'. 

dx  ^  dx  ^  dx 

VI.   —(u^)  =  nu^-^  —  u. 
dx  dx 

yil.    —  (ax'')  =  nax^-\ 
dx 

749.  In  proving  the  formulae  of  §  748,  we  shall  suppose  that, 
when  x-\-h  is  substituted  for  x,  u  is  changed  to  %  +  h',  v  to 
V  +  h",  w  to  w-{-  h"\  etc. 

Proof  of  I. 

(x  +  h)-xl_^ 

^'         J 
That  is,  the  derivative  with  respect  toxofx  itself  is  unity. 

Proof  of  II. 

d  /„  I  «N_   lim  [(u-{-h'  +  a)-(u-^a)l_   lim   pn_  d 
-^u  +  a)-^^^^  I  ]-h:^olh\-d^''' 

by  §  747,  (1). 

That  is,  the  derivative  with  respect  to  x  of  a  function  of  x  plus 
a  constant  equals  the  derivative  of  the  function  of  x. 


By  S  747,         ±.  =  ^% 


THEORY  OF  EQUATIONS  529 


For  example,  ^(So^-5)=-f{3  x^. 

Proof  of  III. 

A 
dx 


/^,,x  _   lim  \~a(u  +  h')  —  au~\  _  lim  [ahn 

That  is,  the  derivative  with  respect  to  x  of  a  constant  times 
a  function  of  x  equals  the  constant  times  the  derivative  of  the 
function  of  x. 

For  example,  —  (3  ar^  =  3  —  (a^). 
dx  dx 

Proof  of  IV. 

-—(ll-\-V-\- IV -{-•'•) 

dx 

^    lim   nu-\-h'-{-v  +  h"'^W-hh"'-\-"')~(u-\-v-{-W-{-'-)~\ 
h  =  0\_  h  J 

^   lim   rh'  +  h"-\-h"'+'''l 
h  =  0\_  h  J 

=;To[l']+."ro[g+."r„[^]+-(^254) 

d       ,    d      ,    d       , 
dx        dx        dx 

That  is,  the  derivative  with  respect  to  x  of  the  sum  of  any 
7iumber  of  functions  of  x  equals  the  sum  of  their  derivatives. 


Proof  of  V. 

Consider  first  the  case  of  two  factors. 

A 

dx 


(^^)=   lim  r(u-\-h')(v  +  h")-uvl 

^   lim   ruh"^(v  +  h")hn 
h  =  0\_  h  J 

=-;ro[l'>;ro[^+'^"]x."ro[g. 

by  §§  254,  257. 


530  ADVANCED  COURSE  IN   ALGEBRA 

As  h  approaches  the  limit  0,  /i"  also  approaches  0,  and  there- 
fore the  limiting  value  oi  v  -\-  h"  is  v. 

Whence,  —  (uv)  =  u-^v-j-v-—u-  (1) 

dx  dx  dx 

Consider  next  the  case  of  three  factors. 

—  (uvw)  =  —  l{uv)  '  w']  =  w—  (uv)  -\-uv-—w,  by  (1) 
dx  (xx  (xx  itx 

/    d      ,       d    \  ,         d 

=:W{U V  -}-V U  ]-\-  UV W. 

\    dx  dx   J  dx 

d       ,         d      ,        d 
=  vw  —  u-{-uw  —  V  -}-uv  —  w. 
dx  dx  dx 

In  like  manner  the  theorem  may  be  proved  for  any  number 
of  factors. 

That  is,  tJie  derivative  with  respect  to  x  of  the  product  of  any 
number  of  functions  of  x  equals  the  sum  of  the  results  obtained  by 
multiplying  the  derivative  of  each  factor  by  all  the  other  factors. 

For  example,  —  [{x  +  1)  x'^  =  {x  +  1)  ^~  (x^  ^x'A(x-\- 1). 
dx  /-IN  dx  dx 

Proof  of  Yl. 

If  we  suppose  v,  w,  •••,  in  V,  to  be  all  equal  to  u,  and  that 
the  number  of  factors  is  w,  we  have  by  V, 

A  (u")  =  it«-i  —  u  +  w"-^  —  u-\ to  n  terms  =  tiw""^  —  u. 

dx  dx  dx  dx 

For  example,  —  [(x'  +  1)']  =  S(x'  +  If  ~  {a?  +  1). 
dx  ax 

Proof  of  Nil. 

By  III,  —  (ax'')  =  a  —  (ic")  =  anx^-^  —  x,  by  VI, 
dx  dx  dx 

=  awa;""^,  by  I. 

That  is,  the  derivative  with  respect  to  x  of  a  constant  times  any 
positive  integral  poiver  of  x  equals  the  constant,  times  the  exponent 
of  the  power,  times  x  raised  to  a  power  whose  exponent  is  less  by  1. 


THEORY  OF  EQUATIONS  531 

For  example,  —  (3  x')  =  12  o^. 

(XX 

Ex.  Find  the  derivative  with  respect  to  x  of 

By  II  and  IV, 

^(2a^-5x'  +  7x-6)  =  —  (2af)-—(5x')  +  —(7x) 
dx  dx  dx      ^  ^      dx 

=  6a;2_10aj  +  7,  by  VII. 


EXERCISE  127 

Find  the  derivative  with  respect  to  a;  of  : 

1.  bx^-\-Tx.  5.   4a;6-7a;*  +  8a:3-ic. 

2.  3  x8  -  a;2  +  3.  6.   x^  -  4  ic*  +  6  ic^  +  7  a;2  _  6. 

3.  aj'i  +  5  ic3  -  12  X  -  4.  7.    2  x^  +  3  x*  -  x^  +  8  x2  +  5  x. 

4.  x5  +  9x4-4x2.  8.    5x6-2x5-4x3+11x2  +  8. 

750.   Successive  Differentiation. 

If  u  is  any  function  of  x,  the  derivative  of  the  derivative  of 
u  is  called  the  Second  Derivative  of  u  with  respect  to  x,  and  is 

d? 
represented  by  —^u- 

(XX 

The  derivative  of  the  second  derivative  of  u  is  called  the 

TJiird  Derivative  of  u  with  respect  to  x,  and  is  represented  by 

d^ 

—  u:  etc. 
dx^    ' 

Ex.   Find  the  successive  derivatives  with  respect  to  x  of 

3a^-9x'-12x  +  2, 

We  have,  —  (3  o^  -  9  aj^  _  12  a;  +-  2)  =  9  a;^  -  18  «  -  12. 
dx 

;^(3aj3- 9a^  -  12  a;  +  2)  =  18»  -  18. 
aou 

-^(3a^-  9a^-  12  a;  +-  2)  =  18. 
dx^ 

—  r3a^-9ar^-12a;+-2)=0:  etc. 
dx*  ^ 


532       ADVANCED  COURSE  IN  ALGEBRA 

It  will  be  understood  hereafter  that  when  we  speak  of  the  derivative  of 
a  function  of  x,  the  first  derivative  is  meant. 

EXERCISE   128 

Find  the  successive  derivatives  with  respect  to  x  of : 

1.  4a;2  +  7x-3.  4.   7 a;*  +  a;^  +  9 x2. 

2.  2x3-ll(c2  +  4.  5.   3a;5  +  2a;4  +  6a:2-5. 
^3.  x^-bx^~2x.  6.   x6  -  4 x5  -  10 ic3  +  13 x. 

751.  Graphical  Representation  of  Derivatives. 
Let  PQ  be  the  graph  of  any  function  of  x,  f(x). 

Let  P  be  any  point  on  the  graph  having  the  abscissa  x,  and 
Q  another  point  having  the  abscissa 
X  -^  h. 

Draw  PM  and  QN  perpendicular  to 
XX' J  and  PP  perpendicular  to  Q]^. 

Then,  PM  represents  f{x),  and  QN 
represents  f{x-\-  Ji). 

Whence,  QB  represents  f(x  +  h)  — 

Then,  the  ratio  /(^  +  ^0 -/(«')  jg  represented  by  S^- 
h  PR 

If  we  denote  the  derivative  off(x)  with  respect  to  x  hy  f'(x), 

fix)  =  ^"-^ \f±±Kj^zlM]^  (§  747).  ^ ' 

Then,  f{x)  is  represented  by  the  limiting  value  of  the  ratio 

■^—  as  h  approaches  the  limit  0 ;  that  is,  as  Q  approaches  P. 
Psi 

Hence,  if  PT  is  tangent  to  the  graph  at  P,  meeting  QN 

/T70 

produced  at  T,  f(x)  is  represented  by  the  ratio  --^• 

PR 
The  latter  ratio  is  the  tangent  of  the  angle  TPB. 

752.  By  application  of  the  principles  of  §  751,  we  can  deter- 
mine the  points  where  any  graph  is  parallel  to  XX'. 

TR 
For,  if  the  graph  is  parallel  to  XX'  at  P,  the  ratio  — — 

-t  Jill 

is-  evidently  equal  to  zero ;  so  that  f\x)  is  zero  at  that  point, 


Y 

o^^^^-~~ 

^ 

T 
R 

^ 

/ 

_ 

-rr 

0 

M      h 

N    ^ 

Y' 

THEORY  OF  EQUATIONS  533 

If  then  we  solve  the  equation  f\x)  =  0,  we  shall  determine 
in  what  points  the  graph  of  f(x)  is  parallel  to  XX'. 
Consider,  for  example,  the  graph  of  §  744. 
Here,/'(a;)  =  3aj2-4a;-2. 
Solving  the  equation  3a^— 4ic— "2  =  0,  we  have 


^_2±V4  +  6_2±V10 

JC  —  — * 

3  3 

Then,  one  point  is  ^-^^^^  to  the  right  of  TY',  and  the 

other  ^^^  ~  ^  to  the  left. 
3 

Let  the  pupil  determine  the  points  where  the  graphs  of  the  first  mem- 
bers of  the  equations  in,  Exs.  1  to  6,  inclusive,  in  Exercise  126,  are  parallel 
to  XX'. 

MULTIPLE   ROOTS 

753.  If  an  equation  has  two  or  more*  roots  equal  to  a,  a  is 
said  to  be  a  Multiple  Root  of  the  equation. 

In  the  above  case,  a  is  called  a  double  root,  triple  root,  quad-  i 
ruple  root,  etc.,  according  as  the  equation  has  two  roots,  three 
roots,  four  roots,  etc.,  equal  to  a. 

754.  Let  the  equation 

jpoaj**  -f-piOJ^-i  4-  •••  +Pn-lX  +Pn  =  0  (1) 

have  m  roots  equal  to  a. 

By  §  717,  the  first  member  can  be  put  in  the  form 

(x^arf(x)',  (2) 

where  f(x)  is  the  product  of  the  factors  corresponding  to  the 
remaining  roots  of  (1),  and  is  therefore  a  rational  and  integral 
expression  of  the  (n  —  m)th  degree  with  respect  to  x. 
By  §  748,  V,  the  derivative  of  (2)  with  respect  to  x  is 

(a,_a)»£/(»,)+/(a,)|-[(»;-a)»]; 
or,  (x  -  aYf'{x)  +  m{x  -  a)'^'f{x),  by  §  748,  VI.  (3) 


534  ADVANCED   COURSE   IN   ALGEBRA 

It  is  evident  that  the  expression  (3)  is  divisible  by  (ic— a)*""^ ; 
and  therefore  the  equation  formed  by  equating  it  to  zero  will 
have  m  —  1  roots  equal  to  a. 

Hence,  if  any  equation  of  the  form  (1)  has  m  roots  equal  to  a, 
the  equation  formed  by  equating  to  zero  the  derivative  of  its  first 
member  will  have  m—  1  roots  equal  to  a. 

It  follows  from  the  above  that,  to  determine  the  existence  of 
multiple  roots  in  an  equation  of  the  form 

we  proceed  as  follows : 

Find  the  H.  C.  F.  of  the  first  member  and  its  derivative. 
If  there  is  no  H.  C.  F.,  there  can  be  no  multiple  roots. 
If  there  is  a  H.  C.  F.,  by  equating  it  to  zero  and  solving  the  re- 
sulting equation,  the  required  roots  may  be  obtained. 

The  number  of  times  that  each  root  occurs  in  the  given  j 
equation  exceeds  by  one  the  number  of  times  that  it  occurs  in  ^ 
the  equation  obtained  by  equating  the  H.  C.  F.  to  zero. 

Ex.   Find  all  the  roots  of 

a;4_6a^  + 12  aj2- 10  aj  + 3  =  0.  (1) 

The  derivative  of  the  first  member  is 

4a^-18a^  +  24a;-10. 

The  H,  C.  F,  of  this  and  the  first  member  of  (1)  is  x'^-2  x+\ 
Solving  the  equation  a^  —  2x-f-l  =  0,  the  roots  are  1  and  1. 
Hence,  the  multiple  roots  of  (1)  are  1,  1,  and  1. 
Subtracting  the  sum  of  1,  1,  and  1  from  6,  the  remaining 
root  is  3. 

EXERCISE   129 
Find  all  the  roots  of  each  of  the  following  equations : 

1.  x8-5a;2  +  3a;+9  =  0.  3.   a;8- 12a;+ 16  =  0. 

2.  a;8+3aj2-9a;-27  =  0.  4.    18a;8  +  16a;2  _  4x  -  4  =  0. 

5.  a;4  +  2  a;8  -  11  a;2  -12  a;  +  36  =  0. 

6.  a;*- 11  a;8  + 36x2. -16  a; -64=0. 


THEORY   OF   EQUATIONS 


535    • 


7.  a:5  -  18  a;3  +  4  x2  +  57  X  +  36  =  0. 

8.  a:4  +  11  x3  +  33  x2  +  6«  -  50  =  0. 

9.  x^  -Sx*  -x^  +  7  x^  -4  =  0. 

10.  x^  +  6x^  +  Ux^  -\-  2x'^  -  12x  -  8  =  0. 

755.   We  will  now  construct  the  graph  of  the  first  member 
of  (1),  §  754. 

The  first  member  can  be  put  in  the 
form  (^  _  i)3(^.  _  3)^  (1) 

This  shows  that  the  graph  cuts  XX' 
at  X  =  1  and  x  =  3. 

Since  {x  —  iy(x  —  3)  is  positive  when 
«  is  <  1,  or  >  3,  and  negative  when  x  ^~~o 
is  between  1  and  3,  the  graph  is  above 
XX '  when  x  is  <  1,  or  >  3,  and  below 
when  x  is  between  1  and  3. 

By  §  748,  V,  the  derivative  of  (1)  is 


(X. 


^  dx^ 


3)  +  {x 


Ux' 


ly 


=  {x-lf  +  {x-3)'^x-lf  =  {x- 
=  {x-l)\A.x-10). 
Equating  this  to  zero,  we  have  a;  =  1  or 


l)2(a;-l  +  3a;-9) 


Then  the  graph  is  parallel  to  XX'  at  x=  1  and  x=-\ 

Li 


752). 


If  any  equation,  with  one  unknown  number,  has  a  multiple  root,  the 
graph  of  its  first  member  is  tangent  to  XX}. 

If  the  root  is  a  triple  root,  the  curve  crosses  the  axis  of  X  at  the  point 
of  tangency,  and  reverses  its  direction  at  that  point. 

If  the  root  is  a  double  root,  the  curve  is  entirely  above,  or  entirely 
below  XX'  at  the  point  of  tangency.     (Compare  §  467.) 

Let  the  pupil  construct  the  graphs  of  the  first  members  of  the  equations 
in  Exs.  1,  2,  3,  4,  9,  and  10,  Exercise  129 ;  finding  all  the  points  where 
the  graphs  are  parallel  to  XX'. 

756.   An  equation  of  the  form  a;"  —  a  =  0  can  have  no  multi- 
ple roots  ;  for  the  derivative  of  a;'*  —  a  is  ?ia;"-^  and  a;"  —  a  and    W 
na;""^  have  no  common  factor  except  unity. 


536  ADVANCED  COURSE  IN  ALGEBRA 

Therefore,  the  n  roots  of  x"'  =  a  are  all  different. 

It  follows  from  the  above  that  every  expression  has  two  dif- 
ferent square  roots,  three  different  cube  roots,  and  in  general  ?i 
different  nth  roots. 

STURM'S  THEOREM 

757.  We  will  now  demonstrate  two  theorems  which  are  used 
in  the  proof  or  application  of  Sturm's  Theorem  (§  758). 

I.  Ifxbe  taken  sufficiently  great,  any  term  of  the  expression 

may  be  made  to  numerically  exceed  the  sum  of  all  the  following 
terms. 

For  the  ratio  of  the  (r  -f-  l)th  term  to  the  sum  of  all  the  fol- 
lowing terms  is 

Pi"^ or  P'- (2^ 

By  taking  x  sufficiently  great,  the  denominator  of  (2)  can  be 
made  numerically  as  small  as  we  please ;  hence,  the  ratio  of 
p^x""''  to  the  sum  of  the  following  terms  can  be  made  numeri- 
cally as  great  as  we  please. 

II.  If  X  be  taken  sufficiently  small,  any  term  of  the  expression 

PoX"" + pix""-^  H f-  p,,_jx  +  _p„ 

may  be  made  to  numerically  exceed  the  sum  of  all  the  preceding 
terms. 

Eor  the  ratio  of  the  (r  +  l)th  term  to  the  sum  of  all  the 
preceding  terms  is 

^i^ ,  or  ^ (3) 

By  taking  x  sufficiently  small,  the  denominator  of  (3)  can  be 
made  numerically  as  small  as  we  please ;  hence,  the  ratio  of 
p^'*'"'  to  the  sum  of  the  preceding  terms  can  be  made  numeri-j 
cally  as  great  as  we  please. 


THEORY  OF  EQUATIONS  537 

758.   Sturm's  Theorem. 

Let  f{x)  =  X-  +  p^x^-^  + . . .  4-  p^_ja;  +  p„  =  0  (1) 

be  an  equation   from  which  the   multiple   roots   have   been 
removed  (§  754). 

Let  fi{x)  denote  the  derivative  of  fix)  with  respect  to  x 
(§  746). 

>    Dividing /(a;)  by/i(a;),  we  shall  obtain  a  quotient  Qi,  with  a 
remainder  of  a  degree  lower  than  that  of  fix). 

Denote  this  remainder,  with  the  sign  of  each  of  its  terms 
changed,  by  fX^),  and  divide  fix)  by  fix),  and  so  on ;  the 
operation  being  precisely  the  same  as  that  of  finding  the 
H.  C.  F.  of  f{x)  and  fix),  except  that  the  signs  of  the  terms 
of  each  remainder  are  to  be  changed,  while  no  other  changes  of 
sign  are  permissible. 

Since,  by  hypothesis,  f{x)  =  0  has  no  multiple  roots,  f(pS)  and 
fix)  have  no  common  divisor  except  unity  (§  754) ;  and  we 
finally  obtain  a  remainder /^(ic)  independent  of  x. 

The  expressions /(x), /i(a7), /2(aj),  "',fn{x)  are  called  Sturm/s 
Functions. 

The  successive  operations  are  represented  as  follows : 

f(^)  =  Qifi{^)-f2(x),  (2) 

f(x)  =  Q,fXx)-f{x),  (3) 

fix)  =  Q,f{x)  -fix),  (4) 


fn-2(x)  =  Qn-lfUx)  -fni^). 

We  may  now  enunciate  JSturm^s  TJieorem. 

If  two  real  numhers,  a  and  b,  are  substituted  in  place  of  x 
in  Sturm^s  Functions,  and  the  signs  noted,  the  difference  between 
the  number  of  variations  of  sign  (§  734)  in  the  first  case  and  that 
in  the  second  equals  the  number  of  real  roots  of  fix)  =  0  lying 
between  a  and  b. 

The  demonstration  of  the  theorem  depends  upon  the  fol- 
lowing principles : 

I.  Two  consecutive  functions  cannot  both  become  0  for  the 
same  value  of  x.  * 


538       ADVANCED  COURSE  IN  ALGEBRA 

For  if,  for  any  value  of  x,  f-^{x)  =  0  and  /^{x)  ==  0,  then,  by  (3), 
f^{x)  =  0 ;  and  since  f^ix)  =  0  and  f.^{x)  =  0,  by  (4),  f^{x)  =  0 ; 
continuing  in  this  way,  we  have  finally  fn(x)  =  0. 

But  by  hypothesis,/,, (a;)  is  independent  of  x,  and  consequently 
cannot  become  0  for  any  value  of  x. 

Hence,  no  two  consecutive  functions  can  become  0  for  the 
same  value  of  x. 

II.  If  any  function,  except  f(x)  and  fn{x),  becomes  0  for  any 
value  of  X,  the  adjacent  functions  have  opposite  signs  for  this  value 
of  X. 

For  if,  for  any  value  of  x,  f{x)  =  0,  then,  by  (3),  we  must 
have/i(a;)  =  —f{x)  for  this  value  of  x. 

Therefore,  fi{x)  and  f^{x)  have  opposite  signs  for  this  value 
of  X ;  for,  by  I,  neither  of.  them  can  equal  zero. 

III.  Let  c  be  a  root  of  the  equation  f^{x)  =  0,  where  f(x)  is 
any  function  except /(cc)  and /„(«.'). 

By  II,  fr-i(x)  and  fr+i(x)  have  opposite  signs  when  x  =  c. 

Now  let  h  be  a  positive  number,  so  taken  that  no  root  of 
fr_i(x)  =  0  or  fr+i{x)  =  0  lies  between  c  —  h  and  c  +  h. 

Then  as  x  changes  from  c  —  h'toc-\-  h,  no  change  of  sign  takes 
place  in/._i(a.")  orf^-^(x)  ;  while  f(x)  reduces  to  zero,  and  changes 
or  retains  its  sign  according  as  the  root  c  occurs  an  odd  or  even 
number  of  times  mf(x)  =  0. 

Therefore,  for  values  of  x  between  c  —  h  and  c,  and  also  for 
values  of  x  between  c  and  c  +  h,  the  three  functions /,._i(a?), /.(a;), 
and/.+i(a;)  present  one  permanence  and  one  variation. 

Hence,  as  x  increases  from  c  —  A  to  c  +  /i,  no  change  occurs  in 
the  number  of  variations  in  the  functions/,._i(a;),/,.(fl?),  a,ndf.+i(x)  ; 
that  is,  no  change  occurs  in  the  number  of  variations  as  x  in- 
creases through  a  root  of  f(x)  —  0. 

IV.  Let  c  be  a  root  of  the  equation  f(x)  —  0 ;  and  let  h  be  a 
positive  number,  so  taken  that  no  root  of  fi{x)  =  0  lies  between 
c—h  and  c  -\-  h. 

Then,  as  x  increases  from  c  —  h  to  c  -{-  h,  no  change  of  sign 
takes  place  in /(a.-) ;  while /(x)  reduces  to  zero,  and  changes  sign. 


THEORY  OF  EQUATIONS  539 

Putting  x  =  c  —  h  in  (1),  we  obtain 

/(c  -  h)  =  (c  -  hf  +  p,(c  -  hy-'  +  . . .  +  P„-i(c  -h)-\-  p„. 

Expanding   by   the   Binomial  Theorem,  and  collecting  the 
terms  involving  like  powers  of  h,  we  have 
/(c  -h)=^  c"  +i>ic"-^  +  •••  +Pn-lC  +i?„ 

-  ;i[nc«-i  +  (^  -  l)i>ic"-' +  - +i>„-i] 
4-  terms  involving  h^,  h^,  ••-,  /i". 
But  since  c  is  a  root  off(x)  =  0,  we  have  by  (1), 
c«  +pic"-i  +  ...  +i>„_ic  +i)„  =  0. 

Also,  it  is  evident  that  the  coefficient  of  —  h  is  the  value  of 
fi(x)  when  c  is  substituted  in  place  of  x ;  therefore, 

f(c  —  h)  =  —  hfi(c)  +  terras  involving  h^,  h^,  -•-,  /i".  (5) 
In  like  manner  it  may  be  shown  that 

/(c  4-  /i)  =  -f-  /i/i(c)  -h  terms  involving  h^,  W,  .-.,  /i".       (6) 

Now  if  h  be  taken  sufficiently  small,  the  signs  of  the  second 
members  of  (5)  and  (6)  will  be  the  same  as  the  signs  of  their 
first  terms,  —  hfi{c)  and  +  lifiic),  respectively  (§  757,  II). 

Hence,  if  li  be  taken  sufficiently  small,  the  sign  of  /(c  —  h) 
will  be  contrary  to  the  sign  of  /i(c),  and  the  sign  of  /(c  +  h) 
the  same  as  the  sign  of  /i(c). 

Therefore,  for  values  of  x  between  c  —  h  and  c,  the  functions 
f{x)  and  fi{x)  present  a  variation,  and  for  values  of  x  between 
c  and  c  -\-h  they  present  a  permanence. 

Hence,  a  variation  is  lost  as  x  increases  through  a  root  of 
/(a;)  =  0. 

We  may  now  demonstrate  Sturm's  Theorem;  for  as  x  in- 
creases from  h  to  a,  supposing  a  algebraically  greater  than  b, 
a  variation  is  lost  each  time  that  x  passes  through  a  root  of 
f(x)  =  0,  and  only  then ;  for  when  x  passes  through  a  root  of 
fj^x)  =  0,  where /(a;)  is  any  function  except  f(x)  and  /„(x),  no 
change  occurs  in  the  number  of  variations. 

Hence,  the  number  of  variations  lost  as  x  increases  from 
6  to  a  equals  the  number  of  real  roots  of  f(x)  =  0  included 
between  a  and  b. 


540  ADVANCED   COURSE  IN   ALGEBRA 

♦ 

759.  It  is  customary,  in  applications  of  Sturm's  Theorem, 
to  speak  of  the  substitution  of  an  indefinitely  great  positive 
number  for  x,  in  an  expression,  as  substituting  +  oo  for  x ;  and 
the  substitution  of  a  negative  number  of  mdefinitely  great 
absolute  value  as  substituting  —  co  for  x. 

The  substitution  of  +00  and  —00  for  x  in  Sturm's  Func- 
tions determines  the  number  of  real  roots  oi  fix)  =  0. 

The  substitution  of  +00  and  0  for  x  determines  the  number 
of  positive  real  roots,  and  the  substitution  of  —  oo  and  0  the 
number  of  negative  real  roots. 

Since  Sturm's  Theorem  determines  the  number  of  real  roots 
of  an  equation,  the  number  of  imaginary  roots  also  becomes 
known  (§  715). 

760.  If  a  sufficiently  great  number  be  substituted  for  x  in 
the  expression 

F(X)  =  Po^*^  +  PiX'^-l  +   •  •  •  +  Pn-lX  +  Pn, 

the  sign  of  the  result  will  be  the  same  as  the  sign  of  its  first 
term,  poX""  (§  757,  I) ;  hence. 

If  -\- CO  be  substituted  for  x  in  F(x),  the  sign  of  the  result  will 
be  the  same  as  the  sign  of  its  first  term. 

If  —CO  be  substituted  for  x  in  F{x),  the  sign  of  the  residt  will 
be  the  saine  as,  or  contrary  to,  the  sign  of  its  first  term,  according 
as  the  degree  of  F(x)  is  even  or  odd. 

761.  We  will  now  consider  an  example. 

Let  it  be  required  to  determine  the  number  and  situation  of 
the  real  roots  of    y^^^)  =  a^l  2a^  -  x'+l  =  0. 
Here,  fi{x)  =  3  x^  —  4:  x  —  1. 

In  the  process  of  finding /^(x),  fs(x),  etc.,  any  positive  numeri- 
cal factor  may  be  omitted  or  introduced  at  pleasure ;  for  the 
sign  of , the  result  is  not  affected  thereby  ;  in  this  way  fractions 
may  be  avoided. 

In  this  case,  we  multiply  f(x)  by  3  to  make  its  first  term 
divisible  by  3  x^. 


THEORY  OF  EQUATIONS  541 

3iB2 


# 

4a;- 

-l)3x'-6a^-    3a;-}-3(aj 
3  a:3  _  4  ^  _       ^ 

-2a;2-    2x  +  3 
3 

-6fl^-   6a;4-9(-2 

-6a:2^    8a;  +  2 

7)_14a;4-7 

(z^2x  +  l 

.•./,(x)  =  2a;-l. 

3a^_    4a;-l 
2 

f .  i^^^HL^-y^^' 

2a;-l)6a;-'-    8a;-2(3aj 
6x'-   Sx 

1: '  ^-' 

-   6x-2 

f,..  M. 

2 

-10a;-4(-5 

-  10  a;  +  5 

-9  .•./8(a;)  =  9... 

Substituting  —  oo  for  ic  in  fix),  fi{x),  fsfx),  and  fs(x)y  the 
signs  are  — ,  +,  — ,  and  +,  respectively  (§  760);  substituting 
0  for  X,  the  signs  are  +,  — ,  — ,  +,  respectively;  and  sub- 
stituting +  00  for  a;,  the  signs  are  all  4-. 

Hence,  the  roots  of  the  equation  are  all  real,  and  two  of 
them  are  positive  and  the  other  negative  (§  759). 

We  now  substitute  various  numbers  to  determine  the  situa- 
tion of  the  roots : 


/W 

/iW 

f2(x) 

Mx) 

a!  =  -oo, 

— 

+ 

— 

+ 

3  variations. 

x  =  -l, 

— 

+ 

— 

+ 

3  variations. 

x  =  0, 

+ 

— 

— 

+ 

2  variations. 

x  =  \, 

— 

—  • 

+ 

+ 

1  variation. 

x  =  2, 

— 

+ 

+ 

+ 

1  variation. 

a;  =  3, 

+ 

+ 

+ 

+ 

no  variation.' 

a;  =  oo, 

+ 

4- 

+ 

+ 

no  variation. 

? 


We  then  know  that  the  equation  has  one  root  between  0  and 
—  1,  one  between  0  and  1,  and  one  between  2  and  3. 


l^ 


542 


ADVANCED   COURSE  IN   ALGEBRA 


Y  ^— ' 


762.  It  will  be  found  useful  to  con- 
struct the  graphs  of  f(x)  and  fi(x),  in 
the  example  of  §  761. 

The  graph  of  f(x)  is  the  ■  curve 
ABC',  cutting  XX'  at  A,  between 
x  =  0  and  x  =  —  1,  at  B,  between  x  =  0 
and  x  =  lf  and  at  C,  between  x  =  2 
and  x  =  S. 

The  graph  of  fi(x)  is  the  dotted  curve  DE',  cutting  XX'  at 
D,  between  x  =  0  and  x  =  —  l,  and  at  E  between  x  =  l  and 
x  =  2. 

To  find  the  abscissas  of  D  and  E,  we  solve  the  equation 

3x'-4.x-l  =  0. 

2  ±  2.64+ 


Then,  x  = 


2±V4  +  3 


1.54+  or  -.21+. 


3  3 

If  we  put  x  =  —  .21,  in  f(x),  the  result  is  positive. 

This  shows  that  D  is  between  O  and  A. 

The  graph  illustrates  in  an  excellent  way  the  truth  of  §  758, 
IV;  that  as  x  increases  from  a  value  just  below  to  a  value  just 
above  a  root  of  f(x)  =  0,  no  change  takes  place  in  the  sign  of 
fi(x),  while  f(x)  reduces  to  zero  and  changes  sign. 

763.  As  X  increases  from  -co  to  +go,  f(x)  and/i(cc)  change 
signs  alternately,  for  they  are  always  unlike  in  sign  just  before 
f(x)  changes  sign  (§  758,  IV) ;  hence,  if  the  roots  of  f(x)  =  0 
and  fi{x)  =  0  are  all  real,  a  root  of  fi{x)  =  0  lies  between  every 
two  adjacent  roots  of  f(x)  =  0. 

That  a  root  of  fi(x)  =  0  lies  between  every  pair  of  adjacent 
roots  of  f(x)  =  0,  is  admirably  shown  in  the  figure  of  §  762. 

764.  We  will  give  one  more  illustrative  example. 
Determine  the  number  and  situation  of  the  real  roots  of 

f(x)  =  4oX^  —  6x  —  5  =  0. 

Here,  fi(x)  =  12  a^  —  6 ;  or,  2  ic^  —  1,  omitting  the  factor  6. 

2x^-l)4:a^-6x-5(2x 

4.a^-2x 

—  4  a?  —  5  .-.  /six)  =  4  a;  +-,  5. 


THEORY  OF  EQUATIONS  543 


2^- 

1 

2 

4a;  +  5)4a;2_ 

2{x 

4a^  + 

5x 

— 

5x- 

2 
4 

. 

— 

20  a;- 

8(- 

-5 

— 

20  a;- 

25 

17         .•./,(;»)  =  -17. 

The  last  step  in  the  division  may  be  omitted ;  for  we  only  need  to 
know  the  sign  of  fsix);  and  it  is  evident  by  inspection,  when  the 
remainder  —  5x  —  2  is  obtained,  that  the  sign  of  fs^x)  will  be  — . 

/W    /.(^)    M^)    M^) 


a;  =  —  00, 

— 

+ 

— 

— 

2  variations. 

x^O, 

— 

— 

+ 

— 

2  variations. 

x  =  l, 

— 

+ 

+ 

— 

2  variations. 

x  =  2, 

+ 

+ 

+ 

— 

1  variation. 

X=(X), 

+ 

+ 

+ 

— 

1  variation. 

Therefore,  the  equation  has  a  real  root  between  1  and  2,  and 
two  imaginary  roots. 

In  substituting  the  numbers,  it  is  best  to  work  from  0  in  either  direc- 
tion, stopping  when  the  number  of  variations  is  the  same  as  has  been 
previously  found  for  +oo  or  —  oo,  as  the  case  may  be. 

EXERCISE   130 

Determine  the  nature  of  the  roots  of  the  following  : 

1.  a;8  +  2  a;2  -  a;  -  1  =  0.  5.   a;*  -  8  a:2  -  8  x  +  1  =  0. 

2.  a;8  +  3  a;  -  5  =  0.  6.   x*  +  2  x^  -  5  a;2  -  10  x  -  3  =  0. 

3.  a:8-5x2-f  2x  +  6=0.  7.   x*  +  3x8  -  3x  +  1  =  0. 

4.  a;8  +  x2-15x-28  =  0.  8.   x*  +  4x8  +  2x2  -  6x  -  7  =  0. 

765.   Continuity. 

A  function  of  x,  f(x),  is  said  to  be  Continuous  at  x  =  a  when 
an  indefinitely  small  change  in  a  produces  an  indefinitely  small 
change  in /(a). 


544  ADVANCED   COURSE  IN  ALGEBRA 

Consider  the  rational  integral  function  of  x 

fix)  =PqX''+P^X''-^  +  ...  ^-Pn-iX  -\-p^. 

Putting  a-\-hin.  place  of  x,  we  have 

/(a  +  h)  =p,(a  +  hy  +i)i(a  +  hy-'  +  ...  +Pn. 
Expanding  by  the  Binomial  Theorem, 

f(a  +  h)  =poa^-\-p,a^-'-  +  -  +Pn-ia  +Pn 

+  terms  involving  h,  h^,  "-,  h"" 
=f(a)  +  terms  involving  h,  h^,  .••,  ^^ 
Then,  f(a  +  h)  —/(a)  =  terms  involving  h,  W,  •••,  /i". 

If  h  be  taken  indefinitely  small,  f(a-\-h)—f{a)  will  be  indefi- 
nitely small ;  for  the  coefficients  of  h,  Jr,  •  • .,  li"-  are  finite. 

Hence,  an  indefinitely  small  change  in  a  produces  an  indefi- 
nitely small  change  in /(a). 

That  is,  a  rational  integral  function  of  x  is  continuous  at  every 
value  ofx. 

It  follows  from  the  above  that  the  graph  of  a  rational  integral 
function  of  ic  is  a  continuous  line,  without  breaks. 

We  have  assumed  this  in  the  figures  of  Chap.  XIV,  §§  465, 
467,  482,  and  483,  and  all  the  figures  of  Chap.  XXXVII. 

766.  We  will  illustrate  a  discontinuous  function  of  a;  by  a 
figure. 

Consider  the  fraction 

^  x  —  1 

Put2/=-^- 
x  —  1 

As  X  increases  from  —oo  to  1,  2/  is 
negative,  and  increases  indefinitely  in     ^ 
absolute  value. 

As  X  increases  from  1  to  oo,  ?/  is  posi- 
tive; and  commencing  with  an  indefi-  y'-fi 
nitely  great  value,  decreases  indefinitely. 

The  graph  consists  of  two  branches,  AB  and  CD ;  and  it  is 
evident  that,  at  ic  =  1,  an  indefinitely  small  change  in  x  pro- 
duces an  indefinitely  great  change  in  y. 


h 


—X 


SOLUTION  OF   HIGHER  EQUATIONS  545 

XXXVIII.  SOLUTION  OF  HIGHER  EQUATIONS 

COMMENSURABLE  ROOTS 

We  shall  use  the  term  commensurable  root,  in  Chap.  XXXVIII,  to 
signify  a  rational  root  expressed  in  Arabic  numerals. 

767.  By  §  722,  an  equation  of  the  nth  degree  in  the  general 
form  (§  712),  with  integral  numerical  coefficients,  cannot  have 
as  a  root  a  rational  fraction  in  its  lowest  terms. 

Therefore,  to  find  all  the  commensurable  roots  of  such  an 
equation,  we  have  only  to  find  all  its  integral  roots. 

Again,  by  §  719,  the  last  term  of  an  equation  of  the  above 
form  is  divisible  by  every  integral  root. 

Hence,  to  find  all  the  commensurable  roots,  we  have  only  to 
ascertain  by  trial  which  integral  divisors  of  the  last  term  are  roots 
of  the  equation. 

The  trial  may  be  made  in  three  ways : 

I.  By  substitution  of  the  supposed  root. 

II.  By  dividing  the  first  member  of  the  equation  by  the 
unknown  number  minus  the  supposed  root  (§  183) ;  in  this 
case,  the  operation  may  be  conveniently  performed  by  Syn- 
thetic Division  (§  733). 

III.  By  Newton's  Method  of  Divisors  (§  769). 

In  the  case  of  small  numbers,  such  as  ±  1,  the  first  method 
may  be  the  most  convenient. 

The  second  has  the  advantage  that,  when  a  root  has  been 
found,  the  process  gives  at  once  the  depressed  equation  (§  716) 
for  obtaining  the  remaining  roots. 

If  the  number  of  divisors  is  large,  the  third  method  will 
be  found  to  involve  the  least  work. 

Considerable  work  may  sometimes  be  saved  by  finding  a 
superior  limit  to  the  positive  roots,  and  an  inferior  limit  to  the 
negative  roots  (§§  739,  740);  for  no  number  need  be  tried 
which  does  not  fall  between  these  limits. 


546       ADVANCED  COURSE  IN  ALGEBRA 

Descartes'  Rule  of  Signs  (§  735)  may  also  be  advantageously 
employed  to  shorten- the  process. 

Any  multiple  root  should  be  removed  (§  754)  before  apply- 
ing either  method. 

Ex.   Find  all  the  roots  of  x^  -  15  a;'^  +  10  a;  +  24  =  0. 

By  Descartes'  Rule,  the  equation  cannot  have  more  than  two 
positive,  nor  more  than  two  negative  roots. 

We  can  write  the  first  member  x^  (x^  —  15)  +  10  ic  +  24 ;  then 
no  root  can  be  as  great  as  4  (§  739). 

Changing  the  sign  of  the  x  term,  the  first  member  becomes 

s;4_i5a;2_-L0a;  +  24. 
We  write  this  in  the  form  (see  Ex.  2,  §  739), 

|\2a^2_45)_^^(^_30)  +  24. 
o  o 

Then  no  negative  root  can  be  as  small  as  —  5  (§  740). 

The  integral  divisors  of  24  which  are  less  than  4,  and  greater 
than  —  5,  are  ±1,  ±2,  ±3,  and  —  4. 

By  substitution,  we  find  that  1  is  not,  and  that  —  1  is,  a  root 
of  the  equation. 

Dividing  the  first  member  by  a;  — 2  and  «  — 3  (§  733),  we 
have 

1  4>0  -15  4-10 +24  [2      ^    1  4-0 -15 +10+24  [3 

2        4-22-24  3 9-18-24 

2-11-12,       0  Rem.  3-6-8,      0  Rem. 

The  work  shows  that  2  and  3  are  roots  of  the  given  equation ; 
and  since  the  equation  cannot  have  more  than  two  positive 
roots,  these  are  the  only  positive  roots. 

The  remaining  root  may^be  found  by  dividing  24  by  the 
product  of  —  1,  2,  and  3  (§  720),  or  by  the  same  process  as 
above. 

Dividing  the  first  member  by  a;  +  2,  a;  +  3,  etc.,  we  have 

1  +0  -15  +10  +24  [^  1  +0  -15  +10  +24  L-3 

-2        4      22  -64  -3        9      18  -84 


-2-11      32-40  -3-6      28-60 


SOLUTION   OF   HIGHER  EQUATIONS  547 

1+0-15+10+24  I  -4 
_  4      16-4-24 
-4160 
The  work  shows  that  the  remaining  root  is  —  4. 

768.  By  §  728,  an  equation  of  the  nth  degree  in  the  general 
form,  with  fractional  coefficients,  may  be  transformed  into 
another  whose  coefficients  are  integral,  that  of  the  first  term 
being  unity. 

The  commensurable  roots  of  the  transformed  equation  may 
then  be  found  as  in  §  767. 

Ex.     Find  all  the  roots  of  4  ar'  - 12  iK^  _^  27  a;  - 19  =  0. 

Dividing  through  by  the  coefficient  of  x^,  we  have 

a^_3a^2^27»_19^0  '    ^. 

4         4 

Proceeding  as  in  §  728,  it  is  evident  by  inspection  that  the 
multiplier  2  will  remove  the  fractional  coefficients ;  thus  the 
transformed  equation  is 

a^  _  2  .  3aj2^  22 .  ?I^- 2« .  1?=  0, , 
4  4 

or,  ar^-6aj2  +  27a;-38  =  0;  (1) 

whose  roots  are  those  of  the  given  equation  multiplied  by  2. 
By  Descartes'  Eule,  equation  (1)  has  no  negative  root. 
The  positive  integral  divisors  of  38  are  1,  2,  19,  and  38. 
Dividing  the  first  member  by  x  —  1,  x  —  2,  etc.,  we  have 

l_6+27-38Ll  1-6+ 27- 38  [_2 

1  -   5       22  2  -    8       38 

-5       22  -16  _4       19         0 

The  work  shows  that  2  is  a  root  of  (1). 

The  remaining  roots  may  now  be  found  by  depressing  the 
equation;  it  is  evident  from  the  right-hand  operation  above 
that  the  depressed  equation  is  x-  —  4  a;  +  19  =  0. 


Solving  this,  a;  =  2  ±  V—  15. 

Thus  the  roots  of  (I)  are  2  and  2  ±  V- 15. 


648  ADVANCED  COURSE   IN   ALGEBRA 

Dividing  by  2,  the  roots  of  the  given  equation  are 


land  1  ± -V— 15. 

z 

]/        769.   Newton's  Method  of  Divisors. 

If  a  is  an  integral  root  of  the  equation 

x^  +  Pia;"-i  H h  p^_2x2  +  p^^-^x  +  p^  =  0, 

where  pi,  •••,  p,i  are  integers,  then 

Transposing,  and  dividing  by  a, 

^  =  -Pn-,-Pn-2a p,a^-'-a--'',  (1) 

a 

from  which  it  is  seen  that  —  must  be  an  integer. 

a 

We  may  write  (1)  in  the  form 

Pn 


a 


-\- Pn-y=  —  Pn-i'^ Pia''-'—  a' 


p 

Kepresenting  ^^  +p„-i  by  g,,_i,  and  dividing  by  a, 

from  which  it  is  seen  that  ^^  must  be  an  integer. 

a 

Proceeding  in  this  way,  it  is  evident  that,  if  a  is  a  root  of 

the   equation,   each   of  -the   numbers   or  -^^,  •••, 

(Id  (1 

or  — >   must  be  an  integer,  and  — +  1  must  equal  0. 

a  a  a  ^ 

We  then  have  the  following  rule : 

Divide  the  last  term  of  the  equation  by  one  of  its  integral 
divisors,  and  to  the  quotient  add  the  coefficient  of  x. 

Divide  the  result  by  the  same  divisor,  and,  if  the  quotient  is  an 
integer,  add  to  it  the  coefficient  of  x^. 

Proceed  in  this  manner  with  each  coefficient  in  succession; 
then,  if  the  divisor  is  a  root  of  the  equation,  each  quotient  will  be 
integral,  and  the  last  quotient  added  to  unity  will  equal  zero. 


SOLUTION  OF   HIGHER  EQUATIONS  549 

If  a  fractional  quotient  is  obtained  at  any  stage,  the  cor- 
responding divisor  is  not  a  root  of  the  equation. 

Ex.    Find  all  the  roots  oi  x^  -  a?  —  7x^  +  x -\- Q>  =  0. 

By  Descartes'  Rule,  the  equation  cannot  have  more  than  two 
positive,  nor  more  than  two  negative  roots. 

The  integral  divisors  of  6  are  ±1,  ±2,  ±3,  and  ±  6. 

By  actual  substitution,  we  find  that  1  and  —  1  are  roots. 

We  will  next  ascertain  if  2  is  a  root ;  a  convenient  arrange- 
ment of  the  work  is  shown  below : 

1  -1  -7  +1  +6[2_ 
2      3 
-5      4 

The  operation  is  carried  out  as  follows : 
Dividing  6  by  2,  gives  3 ;  adding  1,  gives  4. 
Dividing  4  by  2,  gives  2 ;  adding  —  7,  gives  —  5. 
Dividing  —  5  by  2,  the  quotient  is  fractional ;  therefore,  2 
is  not  a  root. 

l-l-7+l+6[3_  l_l_7-M+6  [-2 

-1  -2  __1  __2  -1       3       1-3 

0-3-6       3  0       2-6-2 

In  these  cases,  each  quotient  is  integral,  and  the  last  quotient 
added  to  unity  gives  0  j  therefore,  3  and  —  2  are  roots. 

There  is  no  need  of  trying  +  6  in  this  example,  for  we  know  that  the 
equation  cannot  have  more  than  two  positive  roots. 

EXERCISE  131 

In  each  of  the  following,  find  all  the  commensurable  roots,  and  the 
remaining  roots  when  possible  by  methods  already  given  : 

1.  x8  _   9a;2  +  23ic  -  15  =  0.  4.   «»  +    4 a;2  _    9a;  -  36  =  0. 

2.  x^-    8x2+    5a; +  14  =  0-  5.   Sx"  +  4a;a  -  ISx  +  6  =  0. 

3.  x8  +  12x2 +  44x  + 48  =  0.  6.   4x8  +  16  x2- 7x  -  39  =  0 

7.  X*  +  10  x8  +  36  x2  +  60  X  +  24  =  0. 

8.  x*  -  5  x8  +  20  X  -  16  =  0. 

9.  X*  -  15  x»  +  65  x2  -  105  X  4-  54  =  0. 


550  ADVANCED   COURSE   IN    ALGEBRA 

10.  X*  +  8  x3  +  11  x2  -  32  X  -  60  =  0. 

11.  x*  -  2  a;3  -  17  a;2  +  18  X  +  72  =  0. 

12.  4  x*  -  12  x3  -  9  x2  +  47  X  -  30  =  0. 

13.  6  X*  -  7  x3  -  37  x2  +  8  X  +  12  3::  0. 

14.  x5  +  8  X*  -  7  x3  -  103  x2  +  69  X  +  18  =  0. 

15.  3  X*  +  2  x3  -  18  x2  +  8  =  0. 

16.  x*  +  x3  -  6  x2  +  16  X  -  32  =  0. 

RECIPROCAL  OR  RECURRING  EQUATIONS 

770.  A  Reciprocal  Equation  is  one  such  that  if  any  number 
is  a  root  of  the  equation,  its  reciprocal  is  also  a  root. 

It  follows  from  the  above  that,  if  -  be  substituted  for  a:  in  a 

X 

reciprocal  equation,  the  transformed  equation  will  have  the 
same  roots  as  the  given  equation. 

771.  Let 

a;"  H-Pia;"-1  -{-p^X""-^  -\ \-Pn-2X^  +Pn-lX+Pn  =  0  (1) 

be  a  reciprocal  equation. 

Putting  -  for  x,  the  equation  becomes 

X 

Clearing  of  fractions,  and  reversing  the  order  of  the  terms, 

Pn^""  +  Pn-l^^""'  +  Pn-'P^'"''^  H h  i?2«^  +  Pl^?  +  1  =  0. 

Dividing  through  by  p„, 

Pn  Pn  Pn  Pn  Pn 

By  §  770,  this  equation  has  the  same  roots  as  (1) ;  and  hence 
the  following  relations  must  hold  between  the  coefficients  of 
(1)  and  (2), 

P.  =  ^,P.  =  ^^  ••.P.-.  =  f,P„-.  =  |-',P.=f        (3) 

Pn  Pn  Pn  Pn  Pn 

From  the  last  equation,  p^  =  l\  whence,  p,,  =  ±  1. 

I 


I 


SOLUTION  OF   HIGHER   EQUATIONS  551 

Then  the  equations  (3)  become 

Pl=±Pn-l,    P2=±Pn--2,     '"] 

all  the  upper  signs,  or  all  the  lower  signs,  being  taken  together. 
We  then  have  four  varieties  of  reciprocal  equations : 

1.  Degree  odd,  and  coefficients  of  terms  equally  distant  from 
the  extremes  of  the  first  member  equal  in  absolute  value  and  of 
like  sign ;  as,  ic^  —  2a^  —  2a;-fl  =  0. 

2.  Degree  odd,  and  coefficients  of  terms  equally  distant  from 
the  extremes  of  the  first  member  equal  in  absolute  value  and 
of  opposite  sign ;   as,  3ar^  +  2a5^--i»^-|-a^  —  2a;  —  3  =  0. 

3.  Degree  even,  and  coefficients  of  terms  equally  distant 
from  the  extremes  of  the  first  member  equal  in  absolute  value 
and  of  like  sign ;  as,  cc'^  —  5a^4-6a^  —  5cc  +  l  =  0. 

4.  Degree  even,  and  coefficients  of  terms  equally  distant 
from  the  extremes  of  the  first  member  equal  in  absolute 
value  and  of  opposite  sign,  and  middle  term  wanting;  as, 
2a;«  +  3a^-7a;*  +  7a^-3ic-2  =  0. 

On  account  of  the  properties  stated  above,  reciprocal  equa- 
tions are  also  called  Recurring  Equations. 

772.  Every  reciprocal  equation  of  the  first  variety  may  be 
written  in  the  form 

p^""  +  Pvxf-^  +  p^''-'^  H h  P^  +  Pvc  +  i?o  =  0, 

or,  plx-  +  1)  +Pi{x--'  +x)  +  p^ix""-'  +  a^)  +  -  =  0;        (1) 

or,  pIx-  4- 1)  +  i>ia;(a;"-2  +  1)  +  p^{x--^  + 1)  _^  . . .  =  0  ; 

the  number  of  terms  being  even. 

By  §  142,  since  n  is  odd,  each  of  the  expressions  oj^-f  1, 
rf.n-2  _|_  -^^  g^g^  jg  divisible  by  a;  + 1. 

Therefore,  —  1  is  a  root  of  the  equation. 

Dividing  the  first  member  of  (1)  by  a;  +  l,  the  depressed 
equation  is 

p^ix''-'^  -  a;"-2  +  a;"-3 \-x^-x  +1) 

+  Pi(a;''-^  -  x^-^  +  »"-'* h  «^  —  a^  +  a?) 

+  ^2(35""^  -  «""*  +  a^""* -f.  a;4  _  a^  _^  ^) ^  ...  ^  0. 


55^  ADVANCED   COURSE    IN   ALGEBRA 

Or,      i?oa^'*-^  +  (i>i-Po)aJ"-^+  (7?2-JPi+Po)aJ'*"^+  ... 

which  is  a  reciprocal  equation  of  the  third  variety. 

773.  Every  reciprocal  equation  of  the  second  variety  may 
be  written  in  the  form 

p^""  +  piic''"^  -\-  p.;f(f'~^  4- . . .  _  p^y^  —  p^x — po  =  0, 
or,         p.iof  -  1)  +  p^{x--^  -x)+  pIx^--"  -  x2)  +  . . .  =  0,         (1) 
or,        poix^  -  1)  +  Pixix'^-''  -  1)  +  P'fo\x''-'  -  1)  +  . . .  =  0. 

Since  each  of  the  expressions  x""  —  1,  ic"~^  —  1,  etc.,  iS  divisi- 
ble by  a;  —  1,  +  1  is  a  root  of  the  equation. 

Dividing  the  first  member  of  (1)  by  ic  —  1,  the  depressed 
equation  is 

Poix""-^  +  x""-"  +  x""-^  H h  a^  4-  ic  +  1) 

+ i>2(a5^"^  4-  a?"""*  +  aj""^  +  •••  +  a;*  +  a^  +  aj2)  +  ...  =  0, 
or,  j5oa;^~H  (i>i+i>o)a5'"^+  (P2+i>i+i>o)a3""^+  ••• 

+  (i^2  +  i>i  +  Po)  a^  +  (i>i  +  Po)  a^  +  i?o  =  0 ; 
which  is  a  reciprocal  equation  of  the  third  variety. 

774.  Every  reciprocal  equation  of  the  fourth  variety  may  be 
written  in  the  form 

p,(x^  -  1)  +  p,(x^-^  -  x)  -^p,(x-'^  _  a;2)  4-  ...  =  0,        (1) 
or,         po(^'»-l)+i>ia^(^''"'-l)+P2aj'(i»'^-'-l)  +  -=0; 
the  number  of  terms  being  even  (§  771). 

Since  each  of  the  expressions  x^  —  1,  o?*""^  —  1,  etc.,  is  divisible 
by  x^  —  1,  both  1  and  —  1  are  roots  of  the  equation. 

Dividing  the  first  member  of  (1)  by  x^ —  1,  the  depressed 
equation  is 

Pq{x^-'^  +  x^'"^  +  x""-^  H h  a^^  +  a^  +  1) 

-f-Pi(af*-^  +  x""-^  +  X''-''  H h  a^  +  a^  +  a;) 

-\-p2{xr~'^  +  x""-^  +.a;"-«  H h  a?^  +  a;^  +  aj^)  +  •••  =  0, 

or,  poaJ"-''  +  Pia;"-'  +  (i?2  +i>o)  a;**-"  +  .- 

4-  (i>2  +Po)a^  +PiaJ  +Po  =  0; 
which  is  a  reciprocal  equation  of  the  third  variety 


SOLUTION  OF   HIGHER   EQUATIONS  553 

775.  Every  recijyi'ocal  equation  of  the  third  variety  may  he 
reduced  to  an  equation  of  half  its  degree. 

Let  the  equation  be 

Dividing  through  by  a?*",  the  equation  may  be  written 

4-Pm-/«+^Vp„  =  0.  (1) 

Put         X  -{-  -=z  y. 

X 

Then,  a^4-i  =  ('aj  +  -y-2  =  2/2_2; 

=  2/(2/'-32/)- (2/^-2)  =  2/''-42/^  +  2;  etc. 
In  general, 

an  expression  of  the  rth  degree  with  respect  to  y. 

Substituting  these  values  in  (1),  the  equation  takes  the  form 

776.  It  follows  from  §§  772  to  775  that  any  reciprocal  equa- 
tion  of  the  degree  2  m  + 1,  and  any  reciprocal  equation  of  the 
fourth  variety  of  the  degree  2  m  +  2,  can  be  reduced  to  an 
equation  of  the  mth  degree. 

777.  Ex.   Solve2aj^-5aj^-13a:34-13iB2_^5a._2^0. 
The  equation  being  of  the  second  variety,  one  root  is  1  (§  773). 


554       ADVANCED  COURSE  IN  ALGEBRA 

Dividing  by  a;  —  1,  the  depressed  equation  Is 

a  reciprocal  equation  of  the  third  variety. 

Dividing  by  x\  2(x^  +  i^  _  3  fa?  +  -^  -  16  =  0. 


Putting  a;  +  -  =  2/,  and  a?2  + 1  =/  -  2  (§  775),  we  have 

X  X 

2(f-2)-3y-16  =  0. 
Solving  this  equation,  j^  =  4  or 

Taking  the  first  value,     ic  +  -  =  4,  or  x^  —  4:X  =  —1. 

X 

Whence,  a;  =  2  ±  V3. 

1         5 

Taking  the  second  value,  x-\--  = ,  or  2 a^ -\-5x  =  —  2. 

X         2 

Whence,  a;  =  —  2  or 

'  2 

The  roots  of  the  given  equation  are  1,  —  2, ,  and  2  ±  V3. 

A 

That  2  +  V3  and  2  —  \/3  are  reciprocals  may  be  shown  by  mulMplying 
them  ;  thus,  (2  +  V3)  (2  -  V3)  =  4-3  =  1. 

^  ,       ,     _,      .  EXERCISE  132 

Solve  the  following  : 

1.  4  a;3  +  21  a;2  -f  21  X  +  4  =  0.  3.    cc^  -  5  x2  -  5  x  +  1  =  0. 

2.  ic3  +  4  a;2  -  4  X  -  1  =  0.  4.   6  x*  +  13  ic^  -  13  x  -  6  =  0. 

5.  24  x*  -  10  5c3  -  77  ic2  -  10  X  +  24  =  0. 

6.  x5  +  2  x4  -  6x8  +  5  x2  -  2  X  -  1  =  0. 

7.  5x5  -  56 x*  +  131  x3  +  131  x2  -  56 X  +  5  =  0. 

8.  3  x5  +  4  X*  -  23  x3  -  23  x2  +  4  X  +  3  =  0. 

9.  6  x5  -  7  X*  -  27  x3  +  27  x2  +  7  X  -  6  =  0. 

10.    10  x6  -  19  x5  -  19  x*  +  19  x2  +  19  X  -  10  =  0. 

778.  Binomial  Equations. 

A  Binomial  Equation  is  an  equation  of  the  form  x'^  =  a. 
Binomial  equations  are  also  reciprocal  equations,  and,  in 
certain  cases,  may  be  solved  by  the  method  §  777. 


SOLUTION   OF   HIGHER  EQUATIONS  555 


Putting  X  =  ay,  the  equation  a;**  =  ±  a"  becomes  ?/"  =  ±  1 ; 
which  is  a  form  to  which  every  binomial  equation  may  be 
reduced. 

In  §  460,  methods  were  given  for  the  solution  of  the  bi- 
nomial equations  ^^^  =  ±1,  x^=±l,  and  x^=±l. 

The  forms  x^=  ±1  may  be  solved  by  the  method  of  §  777. 

Binomial  equations  of  any  degree  may  be  solved  by  a  method  involving 
Trigonometry. 

EXERCISE  133 

Solve  the  following : 

1.   x^  =  1.  2.    x^  =  -  1.  3.   x^  =  a^. 

779.   The  Cube  Roots  of  Unity. 

By  Ex.  1,  §  460,  the  roots  of  the  equation  a?^  =  1  are 

1,  =^±^,  and  ^izLV^. 

The  third  root  is  the  square  of  the  second ;  for 

/  _  1  +  V^Ts  Y  _  1  -  2  V^^  -  3  _ -1-V^ 


The  second  root  is  also  the  square  of  the  third.  * 

Hence,  if  the  second  root  be  denoted  by  a>,  the  three  cube 
roots  of  unity  are  1,  w,  and  w^ 

Since  w^  =  1,  they  are  1,  w,  and  —  or  — 

780.  If  the  second  root  be  denoted  by  a,  the  three  roots  are 
a,  am,  and  aw^ ;  for  these  equal  o>,  <o^,  and  a>^  or  1,  respectively. 

In  like  manner,  if  the  third  root  be  denoted  by  a,  the  three 
roots  are  a,  aw,  and  a<o^. 

Hence,  if  either  of  the  cube  roots  of  a  number  be  denoted 
by  a,  the  other  two  roots  are  aa>  and  aw^ 

CUBIC  EQUATIONS 

781.  A  Cubic  Equation  is  an  equation  of  the  third  degree 
(§  113),  containing  but  one  unknown  number. 


556  ADVAK^CED   COtlESE  IN   ALGEBRA 

782.  By  §  732,  the  cubic  equation 

where  pi  is  not  zero,  may  be  transformed  into  another  whose 
second  term  shall  be  wanting  by  substituting  ^  —  ;^'  for  x. 
Hence,  every  cubic  equation  can  be  reduced  to  the  form 
a^  4-  aa?  +  6  =  0. 

783.  Cardan's  Method  for  the  Solution  of  Cubics. 

Let  it  be  required  to  solve  the  equation  ^ -\- ax-\-h~0. 
Putting  X  =  y+^j  the  equation  becomes 

or,  y^j^z^j^{^y^^){y  +  z)^-b  =  (}. 

We  may  give  such  a  value  to  z  that  Syz  +  a  shall  fequal  zero. 

Whence,  z  =  — ^  (1) 

oy 

Then,  2/'  +  5'  +  &  =  Q.  (2) 

Substituting  the  value  of  z  from  (1)  in  (2),  we  have 

f--^^  +  b  =  0,   or   if  +  bf  =  ^'      ■ 
^       21  f  J  ^  y       27 

This  is  an  equation  in  the  quadratic  form  (§  468). 
Solving  by  the  rules  for  quadratics,  we  have 


Theiiby(2),  ^  =  -/-6  =  -|T^|'  +  g-  (4) 

Taking  the  upper  signs  before  the  radicals,  in  (3)  and  (4), 
and  substituting  in  the  equation  x  =  y  -{-z,  we  have 


The  lower  signs  before  the  radicals  give  the  same  value  oix. 
The  other  two  roots  may  be  found  by  depressing  the  given 
equation  (§  716). 


SOLUTION   OF   HIGHER  EQUATIONS  657 

784.     Ex.    Sol  ve  the  equation  a:^  +  3  a^  —  6  cc  +  20  =  0. 
We  first  transform  the  equation  into  another  whose  second 
term  shall  be  wanting. 

Putting  x  =  y-\  =  y-X{^  782),  we  have 
o 

2/3-^2/'  +  32/-l.+  32/^-6?/  +  3-62/  +  6  +  20  =  0, 
or,  2/3_9^_,_28  =  0. 

To   solve  the   latter  equation,   we   substitute  a  =  —  9   and 
6  =  28  in  (5),  §  783. 


Thus,  2/  =  V_  14  +  V196 - 27  +  V_14-V196-27 

=  ^^^  +  V^^^  =  _l_3  =  -4. 
Therefore,  flji=2/  — 1  =  —  5. 

Dividing  the  first  member  of  the  given  equation  by  a;  +  5, 
the  depressed  equation  is 

a;2_2a;.+  4  =  0. 
Solving,  x  =  l±  V^^. 

Thus,  the  roots  of  the  given  equation  are  —  5  and  1  ±  V~3. 

EXERCISE  134 

Solve  the  following : 

1.  x3-245c-72  =  0.  6.    ic3  +  6x2  +  27x-86  =  0. 

2.  ic8  _  12  X  +  16  =  0.  7.    a;3  +  9  x2  +  12  ic  -  144  =  0. 
8.  a;8  +  72  x  +  152  =  0.  8.   ic^  +  x2  -  3  x  +  36  =  0. 

4.  x8  -  12  x2  +  21  X  -  10  =  0.  9.    x3  -  2  x2  -  15  X  +  36  =  0. 

5.  x3  -  3  x2  +  48  X  +  52  =  0.  10.    x^  -  4  x2  +  8  x  -  8  =  0. 

11.   Find  one  root  of  x^  H-  x  —  2  =  0. 
A  cubic  equation  having,  a  commensurable  root  is  solved  more  easily  by 
the  method  of  §  767  than  by  Cardan's  rule. 


785. 


If  h  is  any  one  of  the  cube  roots  of  —  o+vi  '^on^ 

Z        '4       Zi 

and  7c  any  one  of  the  cube  roots  of  "~9~\/x"^97'  ^^®  three 
cube  roots  of  the  first  expression  are  h,  hw,  and  ho)^,  and  the 
three  cube  roots  of  the  second  are  k,  koi,  and  fcw^  (§  780). 


558  ADVANCED  COURSE   IN  ALGEBRA 

This  would  apparently  indicate  that  a;  has  nine  different 

values. 

But  by  (1),  §  783,  yz  —  —  -;  that  is,  the  product  of  the  terms 

o 

whose  sum  is  a  value  of  x  must  be  —  -  • 

o 

Hence,  the  only  possible  values  of  x  are 

h-{-7c,  hu>-{-k(i)^,  and  h<D^  +  ko}', 
for  in  each  of  these  the  product  of  the  terms  is  hk ;  that  is, 

3  /^2         7.2  ^3  ri 

A/ ,  or ;  while  in  any  other  case  the  product  is 

either  — -wor  — -w. 
o  o 

Putting  for  w  and  w^  their  values  (§  779),  the  second  and 
third  values  of  x  become 

Hence,  the  three  values  of  x  are 

,   ,  ,         h-\-k  ,  h  —  k    / — o    ^^A       h  +  k     h  —  k   / — q 
^  +  ^j 2     +~2~  '  2 2~ 

Thus  in  the  example  of  §  784,  h  =  —  \  and  A;  =  —  3. 
Then  the  values  of  y  are  —4,2+  V—  3,  and  2  —  V— 3. 


EXERCISE  135 

Solve  the  following : 

1.     ics  +  12  X  +  12  =  0.  2.   a;3  4. 18  a;  -  6  =  0. 

786.   Discussion  of  the  Solution. 

the  roots  of  a^  +  aa;  +  6  =  0  are 

li^k,  and  J^  ±  ^  V^=r3     (§  785). 


SOLUTION   OF   HIGHER  EQUATIONS  559 

1.  If  a  is  positive,  or  if  a  is  negative  and.  —-  numerically 

less  than  — ,  h  and  k  are  real  and  unequal. 

Therefore,  one  root  is  real,  and  the  other  two  pure  imagi- 
nary or  complex. 

2.  If  a  is  negative,  and  —  numerically  equal  to  — ,  ^  and  k 

are  real  and  equal,  and  ^  —  A;  is  zero. 

Hence,  the  roots  are  all  real,  and  two  of  them  are  equal. 

3.  If  a  is  negative,  and  —  numerically  greater  than  — ,  the 

values  of  h  and  k  involve  pure  imaginary  numbers. 

In  this  case,  h  must  have  some  value  of  the  form  /i'  +  kHy 
where  7i'  and  fc'  are  real  (§  713). 

That  is,       ^  =  ^(- 1  +  Vf+f )  =  ^'  +  ^'*'  (1) 

Then,  ^  =  ^(^-|-.JJ7|)  =  /.'-/^'i.(§424). 

Therefore,     h  +  k=z2h',  and  h-k  =  2  k'i. 

Then  the  three  roots  are  2  h'  and  —h'±  k'iV—  3. 
That  is,  2  h'  and  -h'Tk'  V3. 
Therefore,  the  roots  are  real  and  unequal. 

In  the  above  case,  Cardan's  method  is  of  no  practical  value,  for  since 
there  is  no  method  in  Algebra  for  finding  the  cube  root  of  an  expression 
in  the  form  of  a  rational  expression  plus  a  quadratic  surd  (§  368),  the 
values  of  h  and  k  cannot  be  found ;  in  this  case,  which  is  called  the 
Irreducible  Case.,  Cardan's  method  is  said  to  fail. 

It  is  possible,  in  cases  where  Cardan's  method  fails,  to  find  the  roots 
by  Trigonometry  (see  §  789) ;  but  in  practice  it  is  easier  to  find  them  by 
§  767,  or  by  Horner's  method  (§  794),  according  as  the  equation  has  or 
has  not  a  commensurable  root. 

787.   Consider  the  equation  «^  -f-  aa^  4-  &aj  +  c  =  0. 

Putting  x—y  —  -,  the  equation  becomes 
o  .  . 


560  ADVANCED   COURSE  IN   ALGEBRA 

Or,  f  +  3b^y  +  ^jt:zl^L±Jll^0.  (1) 

Transforming  the  equation  into  another  whose  coefficients 
shall  be  integral,  that  of  the  first  term  being  unity  (§  728),  we 
have  y'^S{3b-a')y-^2d'~9ab  +  27c  =  0, 

whose  roots  are  respectively  3  times  those  of  (1). 

Then  it  follows  from  §  786  that: 

1.  If  3b  —  a^  is  positive,  or  if  3b  — a^  is  negative  and 
4  (3  6  —  a^f  numerically  less  than  (2  a^  —  9  ab-\- 27  cy,  the  given 
cubic  has  one  real  and  two  pure  imaginary  or  complex  roots. 

2.  If  3  6  —  a^  is  negative,  and  4  (3  6  —  a^y  numerically  equal 
to  (2  a^  —  9  ab-\-  27  c)^,  the  roots  are  all  real,  and  two  of  them 
equal. 

3.  If  3  6  —  a^  is  negative,  and  4  (3  &  —  a^)^  numerically  greater 
than  (2  a^  —  9  ab-\-  27  c)^,  the  roots  are  all  real  and  unequal. 

788.  Solution  of  Cubic  Equations  by  Trigonometry  in  Cardan's 
Irreducible  Case. 

To  solve  the  equation  x^  —  ax  —  b  =  0, 

a^      52 
where  a  is  positive,  and  ^  >  -r  •     (Compare  §  786,  3.) 

Putting  x  =  2  m  cos  A,  the  equation  becomes 
8  m^  cos^  A  — 2  am  cos  ^  —  6  =  0; 

or,  4  cos^  ^  —  -^  cos  ^  —  - — -  =  0. 

m-  2  m^ 

It  is  proved  in  Trigonometry  (see  Ex.  16,  p.  35,  Wells^  Com- 
plete Trigonometry)  that 

4  cos^  J.  =  cos  3  ^  +  3  cos  A. 

Whence,  cos  3  ^  +  3  cos  ^  —  — ,  cos  A —  =  0. 

m^  2m^ 

Or,  cos3^  +  (^3-^Vos^l--^  =  0.  (1) 

V       my  2  m^ 


SOLUTION  OF   HIGHER  EQUATIONS  561 

We  may  take  m  so  that    3  —  -^,  =  0 ;  then  m  =  -J- •  (2) 

Then  (1)  becomes  cos  3A  = 

Substituting  in  this  the  value  of  m  from  (2), 

cos3^  =  |^|.  (3) 

Since,  by  hypothesis,  -r  <  t^?  we  have  —  x  -^  <  1. 
4      27  4      a^ 

Taking  the  square  root  of  both  members  of  the  inequality, 
Then,  the  value  of  3^  in  (3)  is  possible,  since  its  cosine 

ig   <^  ;|^^  

Let  z  be  the  least  positive  angle  whose  cosine  equals  j:\—^' 
Then,  one  value  of  3  ^  is  ;$;.  ^ 

Two  other  values  are  2  7r-{-z  and  2  tt  —  2; ;  for  the  cosines  of 
these  angles  are  equal  to  the  cosine  of  z. 

Then,  SA  =  z,  or  2'7r±z; 

and      a;  =  2mcos^  =  2^^cos|  or  2^cosC^±^\ 

where  z  is  given  by  the  equation  cos  z  =  o\/~i* 

An  equation  of  the  third  degree  cannot  have  more  than 
three  different  roots;  so  that  these  are  the  only  values  of  x. 

789.  Ex.     Solve  the  equation 

aj3_4a;-2  =  0. 
i 
Here,  a  =  4,  6  =  2; 

V27 
—  • 

By  logarithms,  log  cos  2;  =  -  (log  27  —  log  64). 


562  ADVANCED   COURSE   IN   ALGEBRA 

log  27=    1.4314 
log  64  =    1.8062 

2)19.6252-20 
log  cos  2=   9.8126-10 
Then,  z  =  4:9°  29.3'. 

Thus,  ^  =  16°  29.8', 

o 

Then,  the  values  of  x  are : 
16 


cos  16°  29.8', 
3 


4 

^15  cos  (120°  + 16°  29.8')  =      aM  cos  136°  29.8' 


-V 


-sin    46°  29.8', 
o 


and     yj—  cos  (120°  -  16°  29.8')  =     yM  cos  103°  30.2' 

=  -^^sin    13°  30.2'. 

log  -^  =  1  (log  16  -  log  3)  =  ^  (1.2041-  .4m)=.3635.  (1) 

log  cos  16°  29.8'  =  9.9817  - 10.  '       (2) 

log  sin  46°  29.8 '  =  9.8606  - 10.  (3) 

log  sin  13°  30.2'  =  9.3683  - 10.  (4) 

Adding /2),  (3),  and  (4)  to  (1),  the  logarithms  of  the  abso- 
lute values  of  x  are 

0.3452,  0.2241,  and  9.7318  - 10. 
The  numbers  corresponding  to  these  are 
2.214,  1.675,  and  .5393. 
Then,  a;  =  2.214,  -1.675,  or  -.5393. 


SOLUTION   OF   HIGHER   EQUATIONS  563 

If  the  given  equation  had  been  r"^  —  4  x  +  2  =  0,  we  should  have  had 

/27 
a  =  4,  &  =  —  2 ;  and  cos  z  would  have  been  —a/ — 

>64 

/27  |27 

If  cos  ^  =  -  ^— ,  COS  (tt  -  S)  =  -  COS  0  =  -y|  — . 

We  should  then  have  found  ir  -z-W  29.3',  and  z  =  130°  30.7'. 
The  three  values  of  x  would  then  have  been  : 

J^  cos  43°  30.2',  and  yj^  cos  (120°  ±  43°  30.2'). 

We  should  have  found  x  =  1.675,  -  2.214,  or  .5393. 

In  any  case  where  a  and  b  are  positive,  z  is  acute,  and  the  equation 
has  one  positive  and  two  negative  roots  ;  if  a  is  positive  and  b  negative, 
z  is  obtuse,  and  the  equation  has  two  positive  and  one  negative  root. 

EXERCISE  136 

Solve  the  following : 

1.  x3  -  4  a;  -  1  =  0.  3.    a;^  +  6  a;2  _  ^  -  1  =  0. 

2.  x3-6x  +  3  =  0.  4.    x3-3x2-2x  +  l=0. 

BIQUADRATIC  EQUATIONS 

790.  A  Biquadratic  Equation  is  an  equation  of  the  fourth 
degree,  containing  but  one  unknown  number. 

Euler's  Method  for  the  Solution  of  Biquadratics. 
By  §  732,  every  biquadratic  equation  can  be  reduced  to  the 
f  oi^ni  x'  +  aa^-^bx  +  c  =  0.  (1) 

Let  x  =  u-{-y-{-z',  then, 

a^  =  u- -\- y^ -^ z^ -{-  2 uy  -{-2 yz  ■^2zu, 
or,  x" -  (u''  +  y'  +  z')  =  2(uy  +  ^JZ  +  zu). 

Squaring  both  members,  we  have 
x^-2x'  (^2  4-  /  +  2")  H-  (u^  +  2/'  +  zy  =  ^(uy-{-yz-\-  zuf 
=  4  (uY  +  i/z^  +  ;2 V)  +  8  uyz  (u-hy  +  z). 
Substituting  x  for  u-^y  -{-  z,  and  transposing, 
x*—2x^ (u^ 4- 2/2 -f  2;^)  —  8  ayzx 

+  (u'  -hy'-h  ^^y  -  4  (uhf  +  7/V  +  z'u"^  =  0. 


664  ADVANCED   COURSE  IN   ALGEBRA 

This  equation  will  be  identical  with  (1)  provided 

b  =  —  Suyz,  (2) 

and  c=:(u'  +  tf-^  z'f  -  4  (uY  +  2/'^'  +  ^'^')- 

Then,  ,^2 _^ / 4- ^2 ^ _ ^,  and  AV=— • 


«'-c 


Also,  uY  +  2/V  +  2:2^2  ^  v_ — ^  ^    ^  —  =  __  =  —j^' 
If,  now,  we  form  the  cubic  equation 

the  values  of  t  will  be  u%  if,  and  ;22  (§  7I8). 
Hence,  if  the  roots  of  the  cubic  equation 

j3  +  «j2  +  «!zii^j_^^0  (3) 

be  I J  m,  and  n,  we  shall  have 

u  =  ±  V^,  y=±  Vm,  and  z  =  ±  \^n. 

Kow  a;  =  1^  +  2/  4-  :2 ;  and  since  each  of  the  numbers  u,  y,  and 
z  has  two  values,  apparently  x  has  eight  values. 

But  by  (2),  the  product  of  the  three  terms  whose  sum  is  a 

value  of  X  must  be 

8 
Hence,  the  only  values  of  x  are,  when  b  is  positive, 

—  V^  —  Vm  —  Vn,  —  V^  +  Vwi  4-  Vn, 
V^  — Vm4-V^,  and  V/+Vm— V?i; 

and  when  b  is  negative, 

vT  +  Vm  +  Vw,  V^  —  Vm  —  V?ij 

—  Vl  H-  Vm  —  Vn,  and  —  V^  —  Vm  +  Vn. 
Equation  (3)  is  called  the  auxiliary  cubic  of  (1). 

791.   Ex.     Solve  the  equation  x*  —  46  or^  —  24  a;  -h  21  =  0. 
Here,  a  =  -46,  &  =  -24,  c  =  21. 


SOLUTION   OF   HIGHER  EQUATIONS  565 

Whence,  ^'7/^  =  127,  and  —  =  9. 

16  '  64 

Then  the  auxiliary  cubic  is  t^  —23t^-{- 127  « —  9  =  0. 

By  the  method  of  §  767,  one  value  of  t  is  9. 

Dividing  the  first  member  by  ^  —  9,  the  depressed  equation 

isf-Ut-^l  =  0.    ,__ 

Solving,  ^=:7±  V49-1  =  7±4V3. 
Proceeding  as  in  §  392,  v^e  have 

V(7±4V3)=V(4±2V12  +  3)  =  2±V3. 
Then  since  b  is  negative,  the  four  values  of  x  are 

3  +  2  +  V3  +  2-V3,  3-2-V3-2+V3, 
-3  +  2  +  V3-2  +  V3,  and  -3-2- V3  +  2- V3. 
^   That  is,  7,  -1,  -3  +  2V3,  and  -3-2V3. 

c  ,        t.    r  11      .  EXERCISE  137 

Solve  the  following : 

1.  x^-mx^-\-S0x  +  SS4:=0. 

2.  x4  -  44  a;2  +  16  X  +  192  =  0. 

3.  a:*  -  40  a;2  +  64  X  +  128  =  0. 

4.  X*  -  54  a:2  -  216  x  -  243  =  0. 

5.  x4  -  22  x2  -  12  X  +  48  =  0. 

6.  X*  +  4  x3  -  4  x2  -  37  X  -  42  =  0. 

792.   Discussion  of  the  Solution. 

The  auxiliary  cubic  of  af*  -j-  ax^  -{- bx  -\- c  =  0  is 

t'  + 1^  +  ^^^t  - 1^  =  0  (§  790).  (1) 

^  lb  o4 

Since  the  last  term  is  essentially  negative,  the  equation 
must  have  either  three  positive,  one  positive  and  two  nega- 
tive, or  one  positive  and  two  pure  imaginary  or  complex  roots 
(§718). 

Transforming  the  equation  into  another  whose  coefficients 
shall  be  integral,  that  of  the  first  term  being  unity  (§  728),  we 
have  t^  +  2at''  +  {ct'-^c)t-b''  =  0,  (2) 

whose  roots  are  respectively  4  times  those  of  (1). 


566  ADVANCED   COURSE   IN   ALGEBRA 

Denoting  2  a,  a^  —  4iC,  and  —  b^  by  a',  b',  and  c',  respectively, 
it  is  necessary,  before  we  can  determine  the  nature  of  the  roots 
of  (2),  to  compute  the  values  of  3  6'  — a'-' and  2a'^  —  9a'b'-{- 
27  c' (§787). 

•Now,  3  6'-a'2  =  3(a2_4c)-4a2=_(a2+l2c), 

and      2a'^-9a'6'  +  27c'  =  16a-^-18a(a2-4c)-2762 

=  -(2a^-72ac  +  2Tb^). 

Then  it  follows  from  §  787  that : 

1.  If  a--\-12  c  is  negative,  or  if  a^-\-12  c  is  positive  and  4  (a^-|- 
12  cy  less  than  (2  a^  —  72  ac  +  27  6-)^,  the  auxiliary  cubic  has 
one  positive  and  two  pure  imaginary  or  complex  roots. 

If  V7=i>j  Vm  =  g  +  rV  — 1,  and  V^  =  g  — rV  — 1,  the 
roots  of  the  biquadratic  are 


—  p±2q  andjp±2rV— 1,  or  p±2q   and  — p±2rV  — 1, 

according  as  b  is  positive  or  negative. 

That  is,  the  biquadratic  has  two  real  and  two  pure  imagi- 
nary or  complex  roots. 

2.   If   o?-\-\2c  is  positive,   and    4 (a^  + 12 cf  is  equal  to 
(2  o?  —  72  ac  +  27  6^)^,  the  cubic  has  two  roots  equal. 
If  Vn  =  Vm,  the  roots  of  the  biquadratic  are 

—  V7  ±  2  Vm,  Vl,  and  V7,  or  VT  ±  2  Vm,  —  VT,  and  —  V^ 

according  as  b  is  positive  or  negative. 

That  is,  the  biquadratic  has  two  roots  equal. 

J  3.  If  a^  H- 12  c  is  positive  and  4(a2  + 12  cf  greater  than 
(2  o?  —  72  ac  +  27  6^^,  the  cubic  has  either  three  positive,  or 
ona  positive  and  two  negative  roots. 

In  the  first  case,  the  roots  of  the  biquadratic  are  all  real ; 
in  the  second  case,  they  are  all  pure  imaginary  or  complex. 

4.   Ifa2+12c  =  0and2a3-72ac  +  2762  =  0,thenc  =  -:^. 

Substituting  from  the  third  equation  in  the  second, 
Sa'-^27b'==0,  ora=-?6*;  whence,  a^  -  4  c  =  ^' =  3  6*. 


^b'== 


SOLUTION   OF   HIGHER  EQUATIONS  567 

In  this  case,  the  auxiliary  cubic  becomes 

'-!*-fe*-£=».»('-f)-»> 

2 

and  each  of  its  roots  is  equal  to  —  • 

The  roots  of  the  biquadratic  are — ,  -^j  -^j  and   -~,  or 

3  6^         6*        6*         ,       6^  ,.  ,     . 

,   —-J7,  — 17>  and  —  — ,    according   as   b    is    positive   or 

z  Z  2d .  Z 

negative;  that  is,  the  biquadratic  has  three  roots  equal. 
5.   If  a^  —  4  c  =  0  and  5  =  0,  the  biquadratic  becomes 

and  its  roots  are  ±^—^  and   ±\/— -• 

That  is,  the  biquadratic  has  two  pairs  of  equal  roots. 

INCOMMENSURABLE  ROOTS 

793.  We  will  now  show  how  to  find  the  approximate  nu- 
merical values  of  those  roots  of  an  equation  which  are  not 
commensurable  (§  767). 

794.  Homer's  Method  of  Approximation. 

Let  it  be  required  to  find  the  approximate  value  of  the  root 
between  3  and  4  of  the  equation 

a:3-3ic2-2icH-5  =  0. 

We  first  transform  the  equation  into  another  whose  roots 
shall  be  respectively  those  of  the  first  diminished  by  3  (§  733).' 
1     _3     _2     +5  13^^. 

ii-\  __?    _9    Zl? 

'-''-^^^  0     -2     -1 

3         9 

3         7 

3 

6 


568  ADVANCED   COURSE   IN   ALGEBRA 

The  transformed  equation  is  y^  -\-  6  y-  -{-  7  y  —  1  =  0.  (1) 

We  know  that  equation  (1)  has  a  root  between  0  and  1. 

If,  then,  we  neglect  the  terms  involving  y^  and  y^,  we  may- 
obtain  an  approximate  value  of  y  by  solving  the  equation 
7?/  — 1=0;  thus,  approximately,  y  =  .l  and  a;  =,3.1. 

Transforming  (1)  into  an  equation  whose  roots  shall  be 
respectively  those  of  (1)  diminished  by  .1,  we  have 

1     +6    +7  -1         \jl 


6    +7 

-1 

.1       .61 
6.1     7.61 

.761 
-   ,239 

a      .62 
6.2    8^3 

.1 
6^ 

The  transformed  equation  is 

23  _^  e_3  ^2  _^  g  23  2  _  .239  =  0.  (2) 

Neglecting  the  z^  and  z^  terms,  we  have,  approximately, 

z  =  -^  =  .02. 
8.23 

Thus,  the  value  of  x  to  two  places  of  decimals  is  3.12. 
Transforming   (2)    into   an   equation  whose  roots  shall   be 
respectively  those  of  (2)  diminished  by  .02,  we  have 

1     +6.3  +8.23         -.239          |^ 

.02  .1264         .167128 

6.32  .8.3564     -.071872 

.02  .1268 

6.34  8.4832 

.02 
6.36 

The  transforuied  equation  is 

u^  +  6.36  u'  +  8.4832  u  -  .071872  =  0. 

Dividing  .071872  by  8.4832,  we  have  .008  suggested  as  the 
fourth  figure  of  the  root. 


SOLUTION   OF   HIGHER  EQUATIONS  569 

Thus,  the  value  of  x  to  three  places  of  decimals  is  3.128. 

The  process  may  be  continued  until  the  value  of  the  root 
has  been  found  to  any  desired  degree  of  precision. 

The  work  is  usually  arranged  in  the  following  form,  the 
coefficients  of  the  successive  transformed  equations  being 
denoted  by  (1),  (2),  (3),  etc. : 


1     -3 

- 

-2 

+  5              13.128 

3 
0 

- 

0 

-2 

(1) 

-6 
-1 

3 
3 

(1)" 

9 

7 

(2) 

.761 
-    .239 

3 

.61 

.167128 

(1)    6 

7.61 

(3) 

-    .071872 

.1 

.62 

6.1 

(2) 

8.23 

.1 

.1264 

6.2 

8.3564 

.1 

.1268 

(2)     6.3 

(3) 

8.4832 

.02 
6.32 

.02 

6.34 

.02 
(3)     6.36 

We  derive  from  the  above  the  following  rule  for  finding  the 
approximate  value  of  a  positive  incommensurable  root : 

Find  by  §§,741  or  742,  or  by  Sturm^s  TJieorem  (§T58),  the 
integral  part  of  the  root.     (Compare  §  743.) 

Transform  the  given  equation  into  another  ivhose  roots  shall  be 
respectively  those  of  the  first  diminished  by  this  integral  part. 

Divide  the  absolute  value  of  the  last  term  of  the  transformed 
equation  by  the  absolute  value  of  the  coefficient  of  the  first  power 
of  the  unknown  number,  and  write  the  approximate  value  of  the 
residt  as  the  next  figure  of  the  root. 


570  ADVANCED   COURSE   IN   ALGEBRA 

Transform  the  last  equation  into  another  whose  roots  shall  be 
respectively  those  of  the  first  diminished  by  the  figure  of  the  root 
last  obtained,  and  divide  as  before  for  the  next  figure  of  the  root; 
and  so  on. 

In  practice,  the  work  may  be  contracted  by  dropping  such  decimal 
figures  from  the  riglit  of  each  column  as  are  not  needed  for  the  required 
degree  of  accuracy. 

In  determining  the  integral  part  of  the  root,  it  will  be  found  convenient 
to  construct  the  graph  of  the  first  member  of  the  given  equation. 

795.  To  find  the  approximate  value  of  a  negative  incommen- 
surable root,  transform  the  equation  into  another  which  shall 
have  the  same  roots  with  contrary  signs  (§  726),  and  find  the 
corresponding  positive  incommensurable  root  of  the  transformed 
equation. 

The  result  with  its  sign  changed  will  be  the  required  negative 
root, 

796.  In  finding  any  particular  root-figure  by  the  method  of 
§  794,  we  are  liable,  especially  in  the  first  part  of  the  process, 
to  get  too  great  a  result ;  the  same  thing  occasionally  happens 
when  extracting  square  or  cube  roots  of  numbers. 

Such  an  error  may  be  discovered  by  observing  the  signs  of 
the  last  two  terms  of  the  next  transformed  equation ;  for  since 
each  root-figure  obtained  as  in  §  794  must  be  positive,  the  last 
two  terms  of  the  transformed  equation  must  be  of  opposite  sign. 

If  this  is  not  the  case,  the  last  root-figure  must  be  diminished 
until  a  result  is  obtained  which  satisfies  this  condition. 

Let  it  be  required,  for  example,  to  find  the  root  between  0 
and  —  1  of  the  equation  oc^-\-4rX^  —  9x  —  5  =  0. 

Changing  the  signs  of  the  x^  term  and  the  independent  term 
(§  726),  we  have  to  find  the  root  between  0  and  1  of  the  equa- 
tion a^-4a^- 9  x-f5  =0  (§  795). 

Dividing  5  by  9,  we  have  .5  suggested  as  the  first  root-figure ; 
but  it  will  be  found  that  in  this  case  the  last  two  terms  of  the 
first  transformed  equation  are  — 12.25  and  —  .375. 


SOLUTION   OF  HIGHER  EQUATIONS  571 

This  shows  that  .5  is  too  great ;  we  then  try  .4,  and  find  that 
the  last  two  terms  of  the  first  transformed  equation  are  of 
opposite  sign. 

The  work  Jf  finding  the  first  three  root-figures  is  shown 
below : 


(1) 


-4 

-    9 

+  5 

|.469 

.4 

-   1.44 

-4.176 

-3.6 

- 10.44 

(1)         .824 

A 

-   1.28 

-    .713064 

-3.2 

(1) 

- 11.72 

(2)         .110936 

.4 

-2.8 

-      .1644 
- 11.8844 

.06 

-      .1608 

-2.74 

(2) 

- 12.0452 

.06 

-2.68 

.06 

(2)  -2.62 
The  required  root  is  —  .469,  to  three  places  of  decimals. 

797.  In  case  too  synall  a  number  is  taken  for  the  root-figure, 
the  number  suggested  for  the  next  root-figure  will  be  greater 
than  .09. 

Let  it  be  required,  for  example,  to  find  the  root  between  0 
and  1  of  the  equation 

a^-2iB2  +  3a;-l  =  0. 

Dividing  1  by  3,  we  have  .3  suggested  as  the  first  root- 
figure.  1 


-2 

+  3 

-1         ^ 

.3 

-    .51 

.747 

-1.7 

2.49 

-    .253 

.3 

-    .42 

-1.4 

2.07 

.3 

-1.1 

The  number  suggested  by  the  next  division  is  greater  than 
.1 J  showing  that  too  small  a  root-figure  has  been  taken. 


572  ADVANCED   COURSE   IN   ALGEBRA 

798.  If  the  coefficient  of  the  first  power  of  the  unknown 
number  in  any  transformed  equation  is  zero,  the  next  figure 
of  the  root  may  be  obtained  by  dividing  the  absolute  value  of 
the  last  term  by  the  absolute  value  of  the  coefficient  of  the  square 
of  the  unknown  number^  and  taking  the  square  root  of  the  result. 

For  if  the  transformed  equation  is  /  +  a/ -{-b  =  0,  it  is  evi- 
dent that,  approximately,  ay^-\-b  =  0,  or  i/  =  •^/ _  _. 

^^      a 

We  proceed  in  a  similar  manner  if  any  number  of  consecu- 
tive terms  immediately  preceding  the  last  term  are  zero. 

Horner's  method  may  be  used  to  find  any  root  of  a  number  approxi- 
mately ;  for  to  find  the  nth  root  of  a  is  the  same  thing  as  to  solve  the 
equation  x"  —  a  =  0. 

799.  If  an  equation  has  two  or  more  roots  which  .have  the 
same  integral  part,  the  first  decimal  root-figure  of  each  must 
be  obtained  by  the  method  of  §§  741  or  742,  or  by  Sturm's 
Theorem. 

If  two  or  more  roots  have  the  same  integral  part,  and  also 
the  same  first  decimal  root-figure,  the  second  decimal  root- 
figure  of  each  must  be  obtained  by  the  method  of  §§  741  or 
742,  or  by  Sturm's  Theorem ;  and  so  on. 

Horner's  method  may  be  used  to  determine  successive  figures  in  the 
integral,  as  well  as  in  the  decimal,  portion  of  the  root. 


If  all  but  one  of  the  roots  of  an  equation  are  known,  the  remaining 
root  may  be  found  by  changing  the  sign  of  the  coefiBcient  of  the  second 
term  of  the  given  equation,  and  subtracting  the  sum  of  the  known  roots 
from  the  result  (§  720). 

EXERCISE  138 

Find  the  root  between  : 

1.  1  and  2,  of  a:^  _  9  ^2  +  23  x  -  16  =  0. 

2.  4  and  5,  of  x3  -  4  x2  -  4  x  +  12  =  0. 

3.  0  and  -^  1,  of  x^  +  8x2  -  9x  -  12  =  0. 

4.  -  2  and  -  3,  of  x8  -  3x2  -  9 X  +  4  =  0. 

5.  3and4,  of  x3-6x2  +  15x-19  =0. 

6.  0  and  1,  of  X*  +  x3  +  2x2  -  X  -  1  =  0. 


SOLUTION  OF   HIGHER  EQUATIONS  573 

7.  2  and  3,  of  x*  -  3  x3  +  4  ic  -  5  =  0. 

8.  -  1  and  -  2,  of  X*  -  2 ic3  -  3x2  +  ic  -  2  =  0. 
Find  all  the  real  roots  of  the  following  : 

9.    x3  +  2  x2  -  X  -  1  =  0.  13.   x8  -  x2  +  2  X  -  1  =  0. 

10.  x8-2x2  -  7x- 1  =0.  14.    x3-x2-15x  +  28  =  0. 

11.  x3-5x2  +  2x  +  6  =  0.  15.   x*-6x2  + llx  +  21  =  0. 

12.  x*  +  2  x3  -  5  =  0.  16.   X*  -  6  xH  6  x2+  8  x  +  1  =  0. 
Find  the  approximate  values  of  the  following : 

17.    v^.  18.    v^21.  19.    v^.  20.    </M. 

800.  We  may  now  give  general  directions  for  finding  the 
real  roots  of  any  equation  of  the  form 

with  integral  numerical  coefficients : 

1.  Determine  by  Descartes'  Eule  (§  735)  limits  to  the  number 
of  positive  and  negative  roots. 

2.  Find  a  superior  limit  to  the  positive  roots,  and  an  infe- 
rior limit  to  the  negative  roots  (§  §  739,  740) . 

3.  Divide  the  first  member  by  x  —  1,  x  —  2,  x  -\- 1,  x  -{-  2, 
etc.,  as  explained  in  §  767. 

In  this  way  all  the  commensurable  roots,  if  any,  will  be 
found,  and  possibly  all  the  incommensurable  roots  may  be 
located. 

4.  If  the  incommensurable  roots  are  not  all  located,  apply 
Sturm's  Theorem ;  observing  that,  if  the  first  member  and  its 
first  derivative  have  a  common  factor,  the  given  equation  has 
multiple  roots  (§  754). 

5.  Approximate  to  the  decimal  portions  of  the  incommen- 
surable roots  by  Horner's  method. 

801.  Newton's  Method  of  Approximation. 

Find  two  consecutive  numbers  a  and  b,  one  greater  and  the 
other  less  than  a  root  of  the  equation  (§  741)  ;  and  suppose  a 
to  be  nearer  the  root  than  b. 

Let  a  +  2/  be  the  actual  value  of  the  root,  and  substitute  iu 
the  given  equation  a-\-y  iov  x. 


574  ADVANCED  COURSE  IN   ALGEBRA 

Then,  y  is  small ;  and  by  neglecting  the  terms  involving  ^/^  2/^ 
etc.,  an  approximate  value  of  y  is  obtained,  which,  when  added  to 
Uf  gives  a',  a  first  approximation  to  the  value  of  x. 

Substitute  in  the  given  equation  o'  +  2;  for  x. 

Then  by  neglecting  the  terms  involving  z',  ^,  etc.,  we  obtain 
an  approximate  value  of  z  which,  when  added  to  a',  gives  a 
second  approximation  to  the  value  of  x. 

The  process  may  be  continued  until  the  root  is  lound  to  any 
desired  degree  of  precision. 

1.  Find  the  root  between  2  and  3,  and  near  2,  of  the 
equation  o^  -  2a;  -  5  =  0. 

Putting  a;  =  2/  +  2,  we  have 

{y  +  2)3-  -  2  (2/  +  2)  -  5  =  0,  or  2/^  +  6  /  +  10  2/  -  1  =  0. 
Then,  approximately,  10  2/  —  1  =  0,  ov  y  =  .1. 
Thus,  2.1  is  a  first  approximation  to  the  value  of  x. 
Putting  in  the  given  equation  x  =  z  -\-  2.1,  we  have 

^3  _^  6,3  ^2  _^  11  23  z  4-  .061  =  0. 
Then,  approximately,  11.23  z  +  .061  =  0,  and  z  =  —  .005+. 
Thus,  2.095  is  a  second  approximation  to  the  value  of  x. 

The  approximate  values  of  y,  z,  etc.,  should  be  obtained  to  one  signifi- 
cant figure. 

2.  Find  the  root  between  —  5  and  —  6,  and  near  —  6,  of  the 
equation  ^  _  0^2  _  25  ^^  4.  81  =  0. 

Putting  a;  =  2/  —  6,  we  have  2/^  —  19  2/^  +  95  y  —  21  =  0. 
Then,  approximately,  952/  —  21  =  0,  and  y  =  .2+. 
Thus,  —  6  4-  .2,  or  —  5.8,  is  a  first  approximation  to  the  value 
of  X. 

Putting  in  the  given  equation  x  =  z  —  5.8,  we  have 

^  -  18.4  z''  4-  87.52  z  -  2.752  =  0. 
Then,  approximately,  87.52  2;  —  2.752  =  0,  and  z=  M-\-. 
Thus,  —  5.8  +  .03,  or  —  5.77,  is  a  second  approximation  to 
the  value  of  x. 


SOLUTION   OF   HIGHER  EQUATIONS  675 

EXERCISE  139 

1.  Find  the  root  between  1  and  2,  and  near  1,  of 

aj3  _  3  x2  -  2  X  +  5  =  0. 

2.  Find  the  root  between  0  and  1,  and  near  1,  of 

x3  +  2  a:2  -  X  -  1  =  0. 

3.  Find  the  root  between  —  2  and  —  3,  and  near  —  2,  of 

4.  Find  the  root  between  —  2  and  —  3,  and  near  —  3,  of 

2c3  _i_  2  a:2  -  5  X  -  7  =  0. 
6.   Find  the  root  between  3  and  4,  and  near  4,  of 

x^-6x^  +  9x-S  =  0. 
6.   Find  the  root  between  —  5  and  —  6,  and  near  —  5,  of 
x3  _  2  a;2  -  23  X  +  70  =  0. 


APPENDIX     .  577 


APPENDIX 

OAUOHT'S  PROOF  THAT  EVERT  EQUATION  HAS  A  ROOT 

802.  We  will  first  prove  that,  if  w  is  a  positive  integer,  each  of  the 
equations  a:"  =  ±  1,  and  x»  =  ±  i 

has  a  root  of  the  form  a  +  bi,  where  a  and  b  are  real  numbers,  either  of 
which  may  be  zero. 

I.  a:'^  =  l. 

It  is  evident  that  1  is  a  root  of  this  equation. 

II.  X**  =  —  1,  where  n  is  odd. 

It  is  evident  that  —  1  is  a  root  of  this  equation. 

III.  x^  =  —  1,  where  n  is  even. 

Let  w  =  2  m,  where  m  is  a  positive  integer ;  then,  a;^"*  =  —  1. 
Extracting  the  square  root  of  each  member,  oj"*  =  ±  i. 
The  latter  forms  are  included  in  the  four  following  cases. 

IV.  X"  =  i,  where  n  is  odd. 

If  m  is  a  positive  integer,  t^'^+i  =  i  (§  411);  hence,  if  n  is  of  the  form 
4m  4- 1,  i  is  a  root  of  the  equation. 

Again,  (-  i)4«M-3  =  _  lim+s  _  _ (_  j)  (§  41 1)  =  i-  hence,  if  n  is  of  the 
form  4  m  +  3,  —  i  is  a  root  of  the  equation. 

V.  x^  =  i,  where  n  is  even. 

Let  n  =  29p,  where  p  is  an  odd  integer ;  then,  x^p  =  i. 

Let  x^^  =  y;  then  yp  =  i,  and,  by  IV,  y  =  i  or  —  i  according  as  j)  is  of 
the  form  4m  +  lor4m  +  3;  that  is,  x^'  =  i  or  —  i. 

The  value  of  x  may  be  obtained  from  this  equation  by  q  successive 
extractions  of  the  square  root ;  and  since  it  has  been  proved  that  the 
square  root  of  a  +  bi,  where  a  and  b  are  real,  can  be  expressed  in  the 
form  a  +  bi  (§  420),  it  follows  that  x  can  be  expressed  in  the  form  a  +  bi. 

VI.  x"^  —  —  I,  where  n  is  odd. 

By  §  411,  (  —  i)4»»+i  =  _  i4m+i  =  —  i;  hence,  if  n  is  of  the  form  4  m  + 1, 
—  i  is  a  root  of  the  equation. 

Again,  i^+^  =  —  i;  hence,  if  n  is  of  the  form  4  m  +  3,  i  is  a  root. 


578  ADVANCED   COURSE   IN  ALGEBRA 

VII.  X"  =  —  ij  where  n  is  even. 

As  in  V,  X  may  be  obtained  in  the  form  a  +  hi. 

803.    We  will  now  consider  the  general  case. 

To  prove  that  the  general  equation  of  the  nth  degree 

X''  +  i?ix'*-i  +  p^^n-l  4. . ..  +  p^_j^  +  Pn  =  0  (1) 

has  a  root  of  the  form  a  +  6i,  where  a  and  6  are  real  numbers. 
Substituting  a  +  &i  for  x  in  (1),  we  have 

(a  +  hiy-{-piia  +  di)**'^  +-  +  Pn-i(a  +  &0  +  P»  =  0. 

Expanding  by  the  Binomial  Theorem,  and  collecting  the  real  and 
imaginary  terms,  we  shall  have  a  result  of  the  form 

Cr+Fi  =  0,  (2) 

where  TJ  and  V  are  real  numbers. 

We  will  now  prove  that  such  real  values  may  be  found  for  a  and  6  as 
will  make  U=0  and  V=0. 

We  will  first  prove  that  such  real  values  may  be  found  for  a  and  h  as 
will  make  U^  +  V^  =  0. 

As  a  and  h  change  in  value,  U  and  V  also  change  ;  and  if  U^  +  V'^ 
cannot  become  zero  for  any  values  of  a  and  6,  it  must  have  some  positive 
real  minimum  value. 

Let  a  and  /S  be  the  values  of  a  and  &,  respectively,  for  which  V"^  +  V^ 
has  this  minimum  value. 

Let  P  +  Qi  be  the  value  of  the  first  member  of  (1)  when  a  +  /3i  is 
substituted  for  x  ;  then  P2  4-  ^2  jg  i\^q  minimum  value  of  U'^  +  V^. 

Writing  a  +  )3i  +  h  in  place  of  x  in  (1),  we  have 


(a  +  /3l  +  ny  +  pi(a  +  ^i  +  ^)'^-i  +  -  +  Pn-\{a  +  ^i  +  /i)  +  p„  =  0. 

Expanding  by  the  Binomial  Theorem,  and  arranging  the  result  in 
ascending  powers  of  h^ 

+  h[_n(ia  4-  /30^-i  +Pi(n  -  l)(a  +  ^iy-^  +  -  +i)n-i] 

+  (terms  involving  /i^,  /^a,  ...,  71'^)=  0.  (3) 

The  first  line  of  (3)  is  equal  to  P  +  Qi. 

The  coefficients  of  some  of  the  powers  of  h  may  be  zero ;  but  they 
cannot  all  be  zero,  since  the  coefficient  of  h^  is  unity. 

Let  Tv"'  be  the  lowest  power  of  h  whose  coefficient  is  not  zero  ;  and 
denote  its  coefficient  by  P  +  8i^  where  B  and  ^S'  are  not  both  zero. 

Then  (3)  becomes         P  +  Qi  +  (P  +  Si)h'^ 

+  (terms  involving  powers  of  h  higher  than  the  mth)  =  0. 


APPENDIX  579 

Let  this  be  denoted  by  P'  +  Q'i  =  0.  (4) 

Now  let  h  =  ct,  where  c  is  a  positive  real  number,  and  t  a  root  of  the 
equation  «"*  =  1  or  «»»  =  —  1. 

By  §  802,  t  is  in  either  case  a  number  of  the  form  a  +  bi. 

Then,  P'  +  Q'i  =  F  +  Qi±(B  +  Si)c^  +  —. 

Whence  (§  419),  F'  =  P±Bc^+  -.., 

and  Q'=Q±Sc^  +  '". 

Therefore,  P'2  +  Q''^  =  I^  +  Q'^±  2(Pi2  +  Q8)c^  +  .... 

That  is,         P'^  +  Q'^-P^-Q'^  =  ±2{PB  +  QS)c^ 

+  (terms  involving  powers  of  c  higher  than  the  mth).  (5) 

If  PB  +  QS  is  not  zero,  c  may  be  taken  so  small  that  the  sign  of  the 
second  member  will  be  the  same  as  that  of  ±  2{PB  +  QS)^^. 

Hence,  if  PB  +  QS  is  positive,  the  sign  of  P  2  +  Q'2  _  p2  _  ^2  may  be 
made  negative  by  taking  «»»  =  -  1 ;  and  if  PB  +  QS  is  negative,  the  sign 
of  P'^  +  Q'^  —  P^  —  Q^  may  be  made  negative  by  taking  i"*  =  +  1. 

Thus,  in  either  case,  P'^  +  Q''^  can,  by  properly  choosing  c  and  t,  be 
made  less  than  P^  +  Q^. 

If  PB  +  QS  =  0,  let  r  =  ±  i  in  (4). 

By  §  802,  t  is  in  either  case  a  number  of  the  form  a  +  hi. 

Then,  P'  +  Q'i  =P+Qi±(B+  Si)ic^  +  — 

=  P+  Qi±  (Bi  -  S)c^  +  — . 

Whence,  P'  =  PtSc^+  ..-, 

and  Q' =  Q  ±  Be"' +  "-. 

Therefore,  P'2+  Q'^  =  P^+  Q^±2{QB- PS)c^+  -.. 

That  is,  P'2  +  Q'2  -  P2  -  §2  =  ±  2  (^i?  -  Pa8')c'«  +  •••• 

Now,  {PB  +  Q8y-\-iQB  -  PSy  =  P^B^  +  Q'^S^  +  Q^B^  +  P2/S2 

=  (P2+^2)(7J2  4.^2). 

And  since,  by  hypothesis,  P2  +  Q2  is  not  zero,  and  B  and  S  are  not 
both  0,  it  follows  that  (PP  +  QS)^  -\-(QB-  PS)^  is  not  zero. 

But  PB-{-  QS  =  0,  and  hence  QB  -  PS  is  not  zero. 

Therefore,  if  c  be  taken  sufficiently  small,  the  sign  of  P'2+  Q''^-P^-  Q^ 
will  be  the  same  as  the  sign  of  ±2  (  ^P  -  P^)c'»  ;  and  we  can  ensure  that 
this  sign  shall  be  negative  by  taking  «»»  =  -  i  when  QB  -  PS  is  positive, 
and  «"»  =  i  when  QB  —  PS  is  negative. 


580      ADVANCED  COURSE  IN  ALGEBRA 

Thus,  by  properly  choosing  h,  P'H  Q'-2  may  be  made  less  than  B^+Q-; 
that  is,  a  value  of  Z/^  +  F^  may  be  obtained  which  is  less  than  P^  +  ^2, 
and  the  latter  is  not  a  minimum  value  of  U"^  +  V'^. 

Hence,  no  positive  real  number  can  be  a  minimum  value  of  U^  +  V'^ ; 
and  therefore  values  of  a  and  b  can  be  found  which  will  make  U^+  V^=0. 

If  U^-\-V2  =  0,  U^  =  -  F2,  and  £/-=  i  Vi. 

Then,  by  §  418,  U=0  and  V=0. 

Hence,  such  real  values  may  be  found  for  a  and  b  as  will  make  U=0 
and  V=0. 

We  will  now  prove  that  the  values  of  a  and  b  which  make  U^-{-V^  =  0 
are  finite. 

The  first  member  of  (1)  may  be  written 


Putting  a  +  bi  in  place  of  x,  we  have 

Consider  the  term 
(a  +  biy     [{a  +  bi)ia-bi)y     (a^  +  b^y^  ^ 

= .  /^o^  r^'  -  ''«'"'^*'  -  ^^^"•^^a'-gft^  +  ...1 

(a2  +  62)rL  |2  ^      J 

=  Ar+  Brh  say. 

'^Lva^  +  W  [2      U2  +  6V      U'  +  W  J 

a  a  b 

Now  if  a  and  6  are  indefinitely  increased  in  absolute  value,  a  +  —  and 

—  +  &  are  indefinitely  increased  in  absolute  value,  for  —  and  —  have  the 
b  a  b 

same  signs  as  a  and  6,  respectively. 

Hence,  if  a  and  b  are  indefinitely  increased  in  absolute  value,  Ar  is 
indefinitely  diminished  in  absolute  value  ;  as  also  is  Br- 

Thus  (6)  may  be  written 

U+  Vi  =  (a  +  biyil  +  A'  +  B'q,  (7) 

where  A'  and  B'  are  indefinitely  diminished  in  absolute  value  when  a  and 
b  are  indefinitely  increased  in  absolute  value. 


APPENDIX  681 

It  a  —  bi  be  substituted  for  x  in  (1),  we  shall  have  a  result  which  may- 
be obtained  from  (7)  by  simply  changing  the  signs  of  the  terms  involv- 
ing i ;  thus,  u_Yi  =  (a-  &0"  [1  +  A'  -  B'f].  (8) 

Multiplying  (7)  and  (8),  U'^  +  V'^=  (a^  +  62)n[(i  +  A'y  ^  B'^l       (9) 

The  second  member  of  (9)  increases  indefinitely  when  a  and  b  are  in- 
definitely increased  in  absolute  value,  for  (a^  +  62^"  increases  indefinitely, 
and  (1  +  A'Y  +  B''^  approaches  the  limit  1. 

Hence,  U^  +  V'^  cannot  be  zero  when  a  and  b,  or  either  of  them,  are 
indefinitely  increased  in  absolute  value,  and  therefore  the  values  of  a  and 
b  which  make  U^  -\-V^  =  0  are  finite. 

804.  The  demonstration  of  §  803  holds  whether  the  coeflBcients  of  the 
terms  in  equation  (1)  are  real,  imaginary,  or  complex. 

It  follows  from  the  above  that  \/— a,  where  n  is  any  even  integer  and  a 
a  positive  real  number,  and  Va  +  bi,  where  n  is  any  positive  integer  and 
a  and  b  any  real  numbers,  can  be  expressed  in  the  form  c  +  di,  where 
c  and  d  are  real  numbers. 

That  is,  any  even  root  of  a  negative  real  number,  or  any  root  of  a 
pure  imaginary  or  complex  number,  can  be  expressed  as  a  pure  imaginary 
or  complex  number.     (Compare  §§  420,  424,  and  439.) 


n 


1  ,./- 


}-. 


m. 


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